Absolute Extrema and Optimization
Mathematics 53
For today
Optimization Problems
maximizing/minimizing a function
f
on a closed and bounded interval
Optimization Problems
maximizing/minimizing a function
f
on a closed and bounded interval
Optimization Problems
maximizing/minimizing a function
f
on a closed and bounded interval
Solving Optimization Problems
1
If possible, draw a
diagram
of the problem corresponding to a general
situation.
2
Assign variables to all quantities involved.
3
Identify the quantity, say
q
, to be maximized or minimized.
4
Formulate an equation involving
q
and other quantities. Express
q
in
terms of a single variable, say
x
. If necessary, use the information
given/relationships between quantities to eliminate some variables.
5
Determine the domain of
q
from the physical restrictions of the
problem, i.e. constraints of
x
.
6
Use appropriate theorems involving absolute extrema to solve the
Solving Optimization Problems
1
If possible, draw a
diagram
of the problem corresponding to a general
situation.
2
Assign
variables
to all quantities involved.
3
Identify the quantity, say
q
, to be maximized or minimized.
4
Formulate an equation involving
q
and other quantities. Express
q
in
terms of a single variable, say
x
. If necessary, use the information
given/relationships between quantities to eliminate some variables.
5
Determine the domain of
q
from the physical restrictions of the
problem, i.e. constraints of
x
.
6
Use appropriate theorems involving absolute extrema to solve the
Solving Optimization Problems
1
If possible, draw a
diagram
of the problem corresponding to a general
situation.
2
Assign
variables
to all quantities involved.
3
Identify the quantity, say
q
, to be
maximized or minimized
.
4
Formulate an equation involving
q
and other quantities. Express
q
in
terms of a single variable, say
x
. If necessary, use the information
given/relationships between quantities to eliminate some variables.
5
Determine the domain of
q
from the physical restrictions of the
problem, i.e. constraints of
x
.
6
Use appropriate theorems involving absolute extrema to solve the
Solving Optimization Problems
1
If possible, draw a
diagram
of the problem corresponding to a general
situation.
2
Assign
variables
to all quantities involved.
3
Identify the quantity, say
q
, to be
maximized or minimized
.
4
Formulate an equation involving
q
and other quantities. Express
q
in
terms of a single variable
, say
x
. If necessary, use the information
given/relationships between quantities to eliminate some variables.
5
Determine the domain of
q
from the physical restrictions of the
problem, i.e. constraints of
x
.
6
Use appropriate theorems involving absolute extrema to solve the
Solving Optimization Problems
1
If possible, draw a
diagram
of the problem corresponding to a general
situation.
2
Assign
variables
to all quantities involved.
3
Identify the quantity, say
q
, to be
maximized or minimized
.
4
Formulate an equation involving
q
and other quantities. Express
q
in
terms of a single variable
, say
x
. If necessary, use the information
given/relationships between quantities to eliminate some variables.
5
Determine the
domain of
q
from the physical restrictions of the
problem, i.e. constraints of
x
.
6
Use appropriate theorems involving absolute extrema to solve the
Solving Optimization Problems
1
If possible, draw a
diagram
of the problem corresponding to a general
situation.
2
Assign
variables
to all quantities involved.
3
Identify the quantity, say
q
, to be
maximized or minimized
.
4
Formulate an equation involving
q
and other quantities. Express
q
in
terms of a single variable
, say
x
. If necessary, use the information
given/relationships between quantities to eliminate some variables.
5
Determine the
domain of
q
from the physical restrictions of the
problem, i.e. constraints of
x
.
6
Use appropriate theorems involving absolute extrema to solve the
Example
Find the number in the interval
[
−
2, 2]
so that the difference of the number
from its square is maximized.
Solution:
(Define variable/s)
Let
x
be the desired number.
(Objective Function)
maximize
f
(
x
)
=
x
2−
x
(Constraint)
x
∈
[
−
2, 2]
≡
I
closed interval!
f
0(
x
)
=
2
x
−
1
C.N.:
12f
(
−
2)
=
6
,
f
(2)
=
2
,
f
(
12)
= −
14Thus
f
attains an absolute maximum on
I
at
x
= −
2
.
Example
Find the number in the interval
[
−
2, 2]
so that the difference of the number
from its square is maximized.
Solution:
(Define variable/s)
Let
x
be the desired number.
(Objective Function)
maximize
f
(
x
)
=
x
2−
x
(Constraint)
x
∈
[
−
2, 2]
≡
I
closed interval!
f
0(
x
)
=
2
x
−
1
C.N.:
12f
(
−
2)
=
6
,
f
(2)
=
2
,
f
(
12)
= −
14Thus
f
attains an absolute maximum on
I
at
x
= −
2
.
Example
Find the number in the interval
[
−
2, 2]
so that the difference of the number
from its square is maximized.
Solution:
(Define variable/s)
Let
x
be the desired number.
(Objective Function)
maximize
f
(
x
)
=
x
2−
x
(Constraint)
x
∈
[
−
2, 2]
≡
I
closed interval!
f
0(
x
)
=
2
x
−
1
C.N.:
12f
(
−
2)
=
6
,
f
(2)
=
2
,
f
(
12)
= −
14Thus
f
attains an absolute maximum on
I
at
x
= −
2
.
Example
Find the number in the interval
[
−
2, 2]
so that the difference of the number
from its square is maximized.
Solution:
(Define variable/s)
Let
x
be the desired number.
(Objective Function)
maximize
f
(
x
)
=
x
2−
x
(Constraint)
x
∈
[
−
2, 2]
≡
I
closed interval!
f
0(
x
)
=
2
x
−
1
C.N.:
12f
(
−
2)
=
6
,
f
(2)
=
2
,
f
(
12)
= −
14Thus
f
attains an absolute maximum on
I
at
x
= −
2
.
Example
Find the number in the interval
[
−
2, 2]
so that the difference of the number
from its square is maximized.
Solution:
(Define variable/s)
Let
x
be the desired number.
(Objective Function)
maximize
f
(
x
)
=
x
2−
x
(Constraint)
x
∈
[
−
2, 2]
≡
I
closed interval!
f
0(
x
)
=
2
x
−
1
C.N.:
12f
(
−
2)
=
6
,
f
(2)
=
2
,
f
(
12)
= −
14Thus
f
attains an absolute maximum on
I
at
x
= −
2
.
Example
Find the number in the interval
[
−
2, 2]
so that the difference of the number
from its square is maximized.
Solution:
(Define variable/s)
Let
x
be the desired number.
(Objective Function)
maximize
f
(
x
)
=
x
2−
x
(Constraint)
x
∈
[
−
2, 2]
≡
I
closed interval!
f
0(
x
)
=
2
x
−
1
C.N.:
12f
(
−
2)
=
6
,
f
(2)
=
2
,
f
(
12)
= −
14Thus
f
attains an absolute maximum on
I
at
x
= −
2
.
Example
Find the number in the interval
[
−
2, 2]
so that the difference of the number
from its square is maximized.
Solution:
(Define variable/s)
Let
x
be the desired number.
(Objective Function)
maximize
f
(
x
)
=
x
2−
x
(Constraint)
x
∈
[
−
2, 2]
≡
I
closed interval!
f
0(
x
)
=
2
x
−
1
C.N.:
12f
(
−
2)
=
6
,
f
(2)
=
2
,
f
(
12)
= −
14Thus
f
attains an absolute maximum on
I
at
x
= −
2
.
Example
Find the number in the interval
[
−
2, 2]
so that the difference of the number
from its square is maximized.
Solution:
(Define variable/s)
Let
x
be the desired number.
(Objective Function)
maximize
f
(
x
)
=
x
2−
x
(Constraint)
x
∈
[
−
2, 2]
≡
I
closed interval!
f
0(
x
)
=
2
x
−
1
C.N.:
12f
(
−
2)
=
6
,
f
(2)
=
2
,
f
(
12)
= −
14Thus
f
attains an absolute maximum on
I
at
x
= −
2
.
Example
Find the number in the interval
[
−
2, 2]
so that the difference of the number
from its square is maximized.
Solution:
(Define variable/s)
Let
x
be the desired number.
(Objective Function)
maximize
f
(
x
)
=
x
2−
x
(Constraint)
x
∈
[
−
2, 2]
≡
I
closed interval!
f
0(
x
)
=
2
x
−
1
C.N.:
12f
(
−
2)
=
6
,
f
(2)
=
2
,
f
(
12)
= −
14Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximize =
constraint: s∈(0, 4.5)
V0(s)= C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximize =
constraint: s∈(0, 4.5)
V0(s)= C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximize =
constraint: s∈(0, 4.5)
V0(s)= C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeVolume=length·width·height
constraint: s∈(0, 4.5)
V0(s)= C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeVolume=(24−2s)·width·height
constraint: s∈(0, 4.5)
V0(s)= C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeVolume=(24−2s)·(9−2s)·height
constraint: s∈(0, 4.5)
V0(s)= C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeVolume=(24−2s)·(9−2s)·s
constraint: s∈(0, 4.5)
V0(s)= C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeV(s)=(24−2s)·(9−2s)·s
constraint: s∈(0, 4.5)
V0(s)= C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeV(s)=2(108s−33s2+2s3)
constraint: s∈(0, 4.5)
V0(s)= C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeV(s)=2(108s−33s2+2s3)
constraint:
s∈(0, 4.5)
V0(s)= C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeV(s)=2(108s−33s2+2s3)
constraint: s∈(0, 4.5)
V0(s)= C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeV(s)=2(108s−33s2+2s3)
constraint: s∈(0, 4.5)
V0(s)=216−132s+12s2
C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeV(s)=2(108s−33s2+2s3)
constraint: s∈(0, 4.5)
V0(s)=4(54−33s+3s2)
C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeV(s)=2(108s−33s2+2s3)
constraint: s∈(0, 4.5)
V0(s)=4(3s−6)(s−9)
C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeV(s)=2(108s−33s2+2s3)
constraint: s∈(0, 4.5)
V0(s)=4(3s−6)(s−9) C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeV(s)=2(108s−33s2+2s3)
constraint: s∈(0, 4.5)
V0(s)=4(3s−6)(s−9) C.N.in(0, 4.5): 2
V00(s)= −132+24s
⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeV(s)=2(108s−33s2+2s3)
constraint: s∈(0, 4.5)
V0(s)=4(3s−6)(s−9) C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeV(s)=2(108s−33s2+2s3)
constraint: s∈(0, 4.5)
V0(s)=4(3s−6)(s−9) C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2
(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeV(s)=2(108s−33s2+2s3)
constraint: s∈(0, 4.5)
V0(s)=4(3s−6)(s−9) C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A rectangular box is to be made from a piece of cardboard 24 inches long and 9 inches wide by cutting out identical squares from the four corners and turning up the sides. Find the volume of the largest rectangular box that can be formed.
Solution:
s
s
s
s s
s s
s
24 in 9 in
s= length of side of squares to be cut out V = Volume of the box formed
maximizeV(s)=2(108s−33s2+2s3)
constraint: s∈(0, 4.5)
V0(s)=4(3s−6)(s−9) C.N.in(0, 4.5): 2
V00(s)= −132+24s ⇒ V00(2)<0
V has a relative maximum ats=2(the only relative extremum on(0, 4.5))
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of the fencing.
highway s2
s1
Solution:
s1= length of side facing the highway
s2= length of other side
C= cost of fencing the lot
minimize
constraint: s1∈(0,∞) :=I
C.N. on I:20
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of the fencing.
highway s2
s1
Solution:
s1= length of side facing the highway
s2= length of other side
C= cost of fencing the lot minimize
constraint: s1∈(0,∞) :=I
C.N. on I:20
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of the fencing.
highway s2
s1
Solution: s1= length of side facing the highway
s2= length of other side
C = cost of fencing the lot
minimize
constraint: s1∈(0,∞) :=I
C.N. on I:20
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of the fencing.
highway s2
s1
Solution: s1= length of side facing the highway
s2= length of other side
C = cost of fencing the lot minimizeC=100s1+40s2
constraint: s1∈(0,∞) :=I
C.N. on I:20
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A1000 square yardrectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of the fencing.
highway s2
s1
Solution: s1= length of side facing the highway
s2= length of other side
C = cost of fencing the lot minimizeC=100s1+40s2
constraint: s1∈(0,∞) :=I
C.N. on I:20
Area: 1000
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of
the fencing. highway
s2
s1
Solution: s1= length of side facing the highway
s2= length of other side
C = cost of fencing the lot minimizeC=100s1+40s2
constraint: s1∈(0,∞) :=I
C.N. on I:20
Area: 1000=s1·s2
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of
the fencing. highway
s2
s1
Solution: s1= length of side facing the highway
s2= length of other side
C = cost of fencing the lot minimizeC=100s1+40s2
constraint: s1∈(0,∞) :=I
C.N. on I:20
Area: 1000=s1·s2
⇒ s2= 1000
s1
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of
the fencing. highway
s2
s1
Solution: s1= length of side facing the highway
s2= length of other side
C = cost of fencing the lot
minimizeC(s1)=100s1+40,000s1
constraint: s1∈(0,∞) :=I
C.N. on I:20
Area: 1000=s1·s2
⇒ s2= 1000
s1
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of
the fencing. highway
s2
s1
Solution: s1= length of side facing the highway
s2= length of other side
C = cost of fencing the lot
minimizeC(s1)=100s1+40,000s1
constraint:
s1∈(0,∞) :=I
C.N. on I:20
Area: 1000=s1·s2
⇒ s2= 1000
s1
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of
the fencing. highway
s2
s1
Solution: s1= length of side facing the highway
s2= length of other side
C = cost of fencing the lot
minimizeC(s1)=100s1+40,000s1
constraint: s1∈(0,∞) :=I
C.N. on I:20
Area: 1000=s1·s2
⇒ s2= 1000
s1
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of the fencing.
highway s2
s1
Solution: s1= length of side facing the highway
s2= length of other side
C = cost of fencing the lot
minimizeC(s1)=100s1+40,000s1
constraint: s1∈(0,∞) :=I
C.N. on I:20
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of the fencing.
highway s2
s1
Solution: s1= length of side facing the highway
s2= length of other side
C = cost of fencing the lot
minimizeC(s1)=100s1+40,000s1
constraint: s1∈(0,∞) :=I
C.N. on I:20
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of the fencing.
highway s2
s1
Solution: s1= length of side facing the highway
s2= length of other side
C = cost of fencing the lot
minimizeC(s1)=100s1+40,000s1
constraint: s1∈(0,∞) :=I
C.N. on I:20
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1
⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of the fencing.
highway s2
s1
Solution: s1= length of side facing the highway
s2= length of other side
C = cost of fencing the lot
minimizeC(s1)=100s1+40,000s1
constraint: s1∈(0,∞) :=I
C.N. on I:20
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of the fencing.
highway s2
s1
Solution: s1= length of side facing the highway
s2= length of other side
C = cost of fencing the lot
minimizeC(s1)=100s1+40,000s1
constraint: s1∈(0,∞) :=I
C.N. on I:20
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the fence is on the highway. The fencing on the highway costs Php80 per yard and the fencing on the other sides costs Php20 per yard. Determine the dimensions of the lot that will minimize the total cost of the fencing.
highway s2
s1
Solution: s1= length of side facing the highway
s2= length of other side
C = cost of fencing the lot
minimizeC(s1)=100s1+40,000s1
constraint: s1∈(0,∞) :=I
C.N. on I:20
C0(s1)=100−40,000s2 1
C00(s1)=2(40,000)s3 1 ⇒ C00(20)>0
⇒REL. MIN. (only rel EXT. on I)
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximize
constraint: r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onIand it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution:
V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximize
constraint: r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onIand it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximize
constraint: r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onIand it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximizeV=πr2h
constraint: r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onIand it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximizeV=πr2h
By similar triangles 6
9
constraint: r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onIand it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximizeV=πr2h
By similar triangles 6
9= 6−r
h
constraint: r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onIand it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximizeV(r)=πr2(9−32r)
By similar triangles 6
9= 6−r
h
constraint: r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onIand it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximizeV(r)=9πr2−32πr3
By similar triangles 6
9= 6−r
h
constraint: r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onIand it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximizeV(r)=9πr2−32πr3
constraint:
r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onIand it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximizeV(r)=9πr2−32πr3
constraint: r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onIand it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximizeV(r)=9πr2−32πr3
constraint: r∈(0, 6)
V0(r)=18πr−92πr2
C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onIand it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximizeV(r)=9πr2−32πr3
constraint: r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onIand it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximizeV(r)=9πr2−32πr3
constraint: r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr
⇒ V00(4)<0
SinceV has only one relative extremum onIand it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximizeV(r)=9πr2−32πr3
constraint: r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onIand it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximizeV(r)=9πr2−32πr3
constraint: r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onI and it is a relative maximum,V has its absolute max at the same point.
∴The inscribed right circular cylinder will have the greatest volume if its dimensions arer=4in and
Example
Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of radius 6 inches and height 9 inches.
Solution: V,h,r= volume, height,radius of inscribed cylinder resp.
6 in
9 in
r
h
maximizeV(r)=9πr2−32πr3
constraint: r∈(0, 6)
V0(r)=18πr−92πr2 C.N.: 4
V00(r)=18π−9πr ⇒ V00(4)<0
SinceV has only one relative extremum onI and it is a relative maximum,V has its absolute max at the same point.
Example
Angelo, who is in a rowboat 2 miles from a straight shoreline notices smoke billowing from his house, which is on the shoreline and 6 miles from the point on the shoreline nearest Angelo. If he can row at 6 miles per hour and run at 10 miles per hour, how should he proceed in order to get to his house at the least amount of time?
Solution:
shoreline Angelo’s house
Starting Point
2 mi
6 mi
c
Example
Angelo, who is in a rowboat 2 miles from a straight shoreline notices smoke billowing from his house, which is on the shoreline and 6 miles from the point on the shoreline nearest Angelo. If he can row at 6 miles per hour and run at 10 miles per hour, how should he proceed in order to get to his house at the least amount of time?
Solution:
shoreline
Angelo’s house
Starting Point
2 mi
6 mi
c
Example
Angelo, who is in a rowboat 2 miles from a straight shoreline notices smoke billowing from his house, which is on the shoreline and 6 miles from the point on the shoreline nearest Angelo. If he can row at 6 miles per hour and run at 10 miles per hour, how should he proceed in order to get to his house at the least amount of time?
Solution:
shoreline Angelo’s house
Starting Point
2 mi
6 mi
c
Example
Angelo, who is in a rowboat 2 miles from a straight shoreline notices smoke billowing from his house, which is on the shoreline and 6 miles from the point on the shoreline nearest Angelo. If he can row at 6 miles per hour and run at 10 miles per hour, how should he proceed in order to get to his house at the least amount of time?
Solution:
shoreline Angelo’s house
Starting Point
2 mi
6 mi
c
Example
Angelo, who is in a rowboat 2 miles from a straight shoreline notices smoke billowing from his house, which is on the shoreline and 6 miles from the point on the shoreline nearest Angelo. If he can row at 6 miles per hour and run at 10 miles per hour, how should he proceed in order to get to his house at the least amount of time?
Solution:
shoreline Angelo’s house
Starting Point
2 mi
6 mi
c
Example
Angelo, who is in a rowboat 2 miles from a straight shoreline notices smoke billowing from his house, which is on the shoreline and 6 miles from the point on the shoreline nearest Angelo. If he can row at 6 miles per hour and run at 10 miles per hour, how should he proceed in order to get to his house at the least amount of time?
Solution:
shoreline Angelo’s house
Starting Point
2 mi
6 mi
c
c = distance bet. house and pt. on the shore from w/c Angelo will start to run
Example
Angelo, who is in a rowboat 2 miles from a straight shoreline notices smoke billowing from his house, which is on the shoreline and 6 miles from the point on the shoreline nearest Angelo. If he can row at 6 miles per hour and run at 10 miles per hour, how should he proceed in order to get to his house at the least amount of time?
Solution:
shoreline Angelo’s house
Starting Point
2 mi
6 mi
c
Example
Angelo, who is in a rowboat 2 miles from a straight shoreline notices smoke billowing from his house, which is on the shoreline and 6 miles from the point on the shoreline nearest Angelo. If he can row at 6 miles per hour and run at 10 miles per hour, how should he proceed in order to get to his house at the least amount of time?
Solution:
shoreline Angelo’s house
Starting Point
2 mi
6 mi
c
c = distance bet. house and pt. on the shore from w/c Angelo will start to run
Example
Angelo, who is in a rowboat 2 miles from a straight shoreline notices smoke billowing from his house, which is on the shoreline and 6 miles from the point on the shoreline nearest Angelo. If he can row at 6 miles per hour and run at 10 miles per hour, how should he proceed in order to get to his house at the least amount of time?
Solution:
shoreline Angelo’s house
Starting Point
2 mi
6 mi
c
c = distance bet. house and pt. on the shore from w/c Angelo will start to run
p
Example
Angelo, who is in a rowboat 2 miles from a straight shoreline notices smoke billowing from his house, which is on the shoreline and 6 miles from the point on the shoreline nearest Angelo. If he can row at 6 miles per hour and run at 10 miles per hour, how should he proceed in order to get to his house at the least amount of time?
Solution:
shoreline Angelo’s house
Starting Point
2 mi
6 mi
c
c = distance bet. house and pt. on the shore from w/c Angelo will start to run
minimize T(c)= p
4+(6−c)2
6 +
c
Example
Angelo, who is in a rowboat 2 miles from a straight shoreline notices smoke billowing from his house, which is on the shoreline and 6 miles from the point on the shoreline nearest Angelo. If he can row at 6 miles per hour and run at 10 miles per hour, how should he proceed in order to get to his house at the least amount of time?
Solution:
shoreline Angelo’s house
Starting Point
2 mi
6 mi
c
c = distance bet. house and pt. on the shore from w/c Angelo will start to run
p
minimizeT(c)= p
4+(6−c)2
6 +
c
10 on [0, 6]
T0(c) = 1 6·
2(6−c)(−1)
2p4+(6−c)2+ 1 10
= 1 6·
(c−6) p
4−(6−c)2+ 1 10
= 5c−30+3 p
4+(6−c)2
30p4+(6−c)2
C.N.: 92
T(0)= p
40 6
T³92´=5260
⇒absolute minimum
T(6)=1415
5c−30+3p4+(6−c)2
30p4+(6−c)2 =0
⇒ 5c−30+3p4+(6−c)2=0
⇒ 5c−30=3p4+(6−c)2
⇒ 25c2−300c+900=9(4+(6−c)2)
⇒ 25c2−300c+900=360−108c+9c2
⇒ 16c2−192c+540=0
⇒ 4c2−48c+135=0
⇒ (2c−9)(2c−15)=0
minimizeT(c)= p
4+(6−c)2
6 +
c
10 on [0, 6]
T0(c) = 1 6·
2(6−c)(−1)
2p4+(6−c)2+ 1 10
= 1 6·
(c−6) p
4−(6−c)2+ 1 10
= 5c−30+3 p
4+(6−c)2
30p4+(6−c)2
C.N.: 92
T(0)= p
40 6
T³92´=5260
⇒absolute minimum
T(6)=1415
5c−30+3p4+(6−c)2
30p4+(6−c)2 =0
⇒ 5c−30+3p4+(6−c)2=0
⇒ 5c−30=3p4+(6−c)2
⇒ 25c2−300c+900=9(4+(6−c)2)
⇒ 25c2−300c+900=360−108c+9c2
⇒ 16c2−192c+540=0
⇒ 4c2−48c+135=0
⇒ (2c−9)(2c−15)=0
minimizeT(c)= p
4+(6−c)2
6 +
c
10 on [0, 6]
T0(c) = 1 6·
2(6−c)(−1)
2p4+(6−c)2+ 1 10
= 1 6·
(c−6) p
4−(6−c)2+ 1 10
= 5c−30+3 p
4+(6−c)2
30p4+(6−c)2
C.N.: 92
T(0)= p
40 6
T³92´=5260
⇒absolute minimum
T(6)=1415
5c−30+3p4+(6−c)2
30p4+(6−c)2 =0
⇒ 5c−30+3p4+(6−c)2=0
⇒ 5c−30=3p4+(6−c)2
⇒ 25c2−300c+900=9(4+(6−c)2)
⇒ 25c2−300c+900=360−108c+9c2
⇒ 16c2−192c+540=0
⇒ 4c2−48c+135=0
⇒ (2c−9)(2c−15)=0
minimizeT(c)= p
4+(6−c)2
6 +
c
10 on [0, 6]
T0(c) = 1 6·
2(6−c)(−1)
2p4+(6−c)2+ 1 10
= 1 6·
(c−6) p
4−(6−c)2+ 1 10
= 5c−30+3 p
4+(6−c)2
30p4+(6−c)2
C.N.: 92
T(0)= p
40 6
T³92´=5260
⇒absolute minimum
T(6)=1415
5c−30+3p4+(6−c)2
30p4+(6−c)2 =0
⇒ 5c−30+3p4+(6−c)2=0
⇒ 5c−30=3p4+(6−c)2
⇒ 25c2−300c+900=9(4+(6−c)2)
⇒ 25c2−300c+900=360−108c+9c2
⇒ 16c2−192c+540=0
⇒ 4c2−48c+135=0
⇒ (2c−9)(2c−15)=0
minimizeT(c)= p
4+(6−c)2
6 +
c
10 on [0, 6]
T0(c) = 1 6·
2(6−c)(−1)
2p4+(6−c)2+ 1 10
= 1 6·
(c−6) p
4−(6−c)2+ 1 10
= 5c−30+3 p
4+(6−c)2
30p4+(6−c)2
C.N.: 92
T(0)= p
40 6
T³92´=5260
⇒absolute minimum
T(6)=1415
5c−30+3p4+(6−c)2
30p4+(6−c)2 =0
⇒ 5c−30+3p4+(6−c)2=0
⇒ 5c−30=3p4+(6−c)2
⇒ 25c2−300c+900=9(4+(6−c)2)
⇒ 25c2−300c+900=360−108c+9c2
⇒ 16c2−192c+540=0
⇒ 4c2−48c+135=0
⇒ (2c−9)(2c−15)=0
minimizeT(c)= p
4+(6−c)2
6 +
c
10 on [0, 6]
T0(c) = 1 6·
2(6−c)(−1)
2p4+(6−c)2+ 1 10
= 1 6·
(c−6) p
4−(6−c)2+ 1 10
= 5c−30+3 p
4+(6−c)2
30p4+(6−c)2
C.N.: 92
T(0)= p
40 6
T³92´=5260
⇒absolute minimum
T(6)=1415
5c−30+3p4+(6−c)2
30p4+(6−c)2 =0
⇒ 5c−30+3p4+(6−c)2=0
⇒ 5c−30=3p4+(6−c)2
⇒ 25c2−300c+900=9(4+(6−c)2)
⇒ 25c2−300c+900=360−108c+9c2
⇒ 16c2−192c+540=0
⇒ 4c2−48c+135=0
⇒ (2c−9)(2c−15)=0
minimizeT(c)= p
4+(6−c)2
6 +
c
10 on [0, 6]
T0(c) = 1 6·
2(6−c)(−1)
2p4+(6−c)2+ 1 10
= 1 6·
(c−6) p
4−(6−c)2+ 1 10
= 5c−30+3 p
4+(6−c)2
30p4+(6−c)2
C.N.: 92
T(0)= p
40 6
T³92´=5260
⇒absolute minimum
T(6)=1415
5c−30+3p4+(6−c)2
30p4+(6−c)2 =0
⇒ 5c−30+3p4+(6−c)2=0
⇒ 5c−30=3p4+(6−c)2
⇒ 25c2−300c+900=9(4+(6−c)2)
⇒ 25c2−300c+900=360−108c+9c2
⇒ 16c2−192c+540=0
⇒ 4c2−48c+135=0
⇒ (2c−9)(2c−15)=0
minimizeT(c)= p
4+(6−c)2
6 +
c
10 on [0, 6]
T0(c) = 1 6·
2(6−c)(−1)
2p4+(6−c)2+ 1 10
= 1 6·
(c−6) p
4−(6−c)2+ 1 10
= 5c−30+3 p
4+(6−c)2
30p4+(6−c)2
C.N.: 92
T(0)= p
40 6
T³92´=5260
⇒absolute minimum
T(6)=1415
5c−30+3p4+(6−c)2
30p4+(6−c)2 =0
⇒ 5c−30+3p4
