2.3 Maximum and Minimum Applications

2.3 Maximum and Minimum Applications

Maximizing (or minimizing) is an important technique used in various fields of study. In business, it is important to know how to find the maximum profit and how to minimize loss. In physics, it is important to find values such as the

maximum speed of an object, or the minimum distance between two moving bodies. In Calculus, maximization is done by analyzing derivatives of functions, a topic beyond our reach now. However, we can look at these problems numerically, using graphing calculators and the properties of the graph of an equation to find the maximum or minimum values. This section will deal with several different types of these problems.

We start off with max/min problems of a type of that should be familiar to you. These are the ones involving a quadratic equation: y=ax2 +bx+c.

Example 1 (Quadratic equations)

The height above ground of an object at time t (in seconds) traveling vertically and subject only to the pull of gravity is given by the equation h , where h o ot h v t + + − = ₁₆ 2

o is the initial height (in feet) and vo is the initial velocity (in feet per

second). A bullet fired upward from ground level has an initial velocity of 480 feet per second. How long does it take to reach its maximum height? How high does it go?

Solution: Here are two ways to solve this problem.

I. Since the equation h=−16t2 +480t

(

)

A. The points on the graph are the ordered pairs (t, h), where h is the height at the time t.

B. The graph is a parabola which will have a vertex. C. Since a = –16 is negative, the graph opens downward.

D. Since B and C are true, the vertex must be the highest point on the graph (maximum).

Both questions will be answered if we find the ordered pair for the vertex. This can be done by completing the square or by using the fact that

a b 2 t = − at the vertex. From this formula we find 15

32 480 ) 16 ( 2 480 ₌ − − = − − = t . This

tells us that it takes 15 seconds for the bullet to reach its maximum height. The maximum height is then found by finding h when t is 15. So

. The maximum height is 3600 feet. ƒ 3600 ) 15 ( 480 ) 15 ( 16 2+ = − = h

II. We can also solve this by graphing the equation and finding the maximum using a calculator. Here are the steps to accomplish this task.

t t

h₌₋16 2 ₊480

A. Graph Y on the graphing calculator. Remember that X is t and Y x x 480 16 2 1 =− + 1 is h.

B. We must adjust the viewing window to see all the important features of the graph.

On CD: • Graphing

Polynomials

C. Setting the window as below we get the following graph.

On CD: • Finding the

Max and Min of

Polynomials

D. To find the maximum (which is the vertex), we use the maximum finder and get x = 15 and y = 3600.

(Note: When you use a calculator to solve a problem you are introducing round-off errors. The resulting answers are usually not exact.) It takes 15 seconds for the bullet to reach a maximum height of 3600 feet. ƒ

Now we will look at some examples that deal with cubic and other equations. Note that at this time, we can solve these problems only by using the graphing calculator. Algebraic approaches are introduced in Calculus.

35 x W L x x 20

Example 2 (Maximizing volume)

An open-top box is to be made from a sheet of cardboard that measures 20 inches by 35 inches by cutting squares of equal size from each of the corners and then folding the flaps up. What size squares should be removed to maximize the volume of the box?

Solution:

h blem involves finding the volume of a box so we need to use the rmula for the volume: V = LWH.

A. What are the length (L), width (W), and height (H) for our box? Well, the length of the original sheet of cardboard is 35 inches. In order to make the box, we will be removing a square of length x from each corner and folding along the dotted lines. After doing so, we have that L = 35 – x –x = 35 – 2x. The same process takes place for the width so we have W = 20 – 2x. The height of the box is determined by the dimensions of the squares that were removed so H = x. Then

T is pro fo x x x V =(35−2 )(20−2 ) . B. The volume must now be maximized. To do this, we graph the

equation in an appropriate window. Note that Y1 is the volume

C. maximum volume for t

which is the x-value.

at

[Co

vie r to construct a box in this fashion we ust remove less than 10 inches of material from each corner. If we

moved mo wouldn’t have a box! Finding

sable value x will then require that you look at e tabl

. The equa e is not a quadratic so the

lgebraic tec ple 1 does not apply.]

Use the maximum finder to find the he

box. When using the maximum finder, we get that the maximum volume is approximately 1296.584, which is the y-value. The size of the squares that are removed is approximately 4.098,

Since we were asked to find the size of the squares, we have th they are 4.098 inches by 4.098 inches. ƒ

mments: 1. How did we know to let x go from 0 to 10 in the wing window? In orde

m

re re than 10 inches then we s for Ymin and Yma e, or “trace” along the graph.

tion that we have for the volum hnique used in exam

u th 2 a

Example 3 (Maximizing area)

Given the equation

with vertices at the um area.

Solution:

2

8 x

y= − , find the positive value of x for which the triangle points (0, 0), (x, 0), and (x, y) has maxim

(0,0) (x, 0) (x, y)

r the area of a triangle A= 1bh We need to use the formula fo

? In this case the base is

A. What are the base b and the height h for our problem x and the height is y. So the area is A

2 1 = ) is on the graph of y =8 x− 2, the fo

xy . Since x, y

( rmula for the area becomes

(

)

B. Now we graph the equation in an appropriate window. We choose the values for x to go from -2 to 5 since the actual values for x must be between 0 and 8 . Note that Y here is the area A.

C. Use the calculator to find that the maximum y-value for this graph. Remember, the y-value represents the area.

1

So an x-value of approximately 1.633 will give you a triangle with

[Comments equation was a quadratic, the

maximum area of approximately 4.35. ƒ : Even though the original

equation that we found for the area was not quadratic. Since is not quadratic, the algebraic form

(

)

2 1

x

x − ula for the “vertex”

Example 4 (

Find the minim ).

Solution:

Minimizing distance)

um distance from the graph of y ₌x3 ₋3x _{and the point (1, 1}

We need to find the formula for the distance from the point (1, 1) to all points on the graph.

A. The points on the graph are the ordered-pairs (x, y) which satisfy . Then a typical point on the graph is

x x

y = 3−3

(

)

B. To find the distance d between the points (1, 1) and

(

)

(

)

(

)

d . This is the quantity we want to

minimize.

C. Graph the equation in the appropriate window, where Y1 is the

distance d given by the formula above. Then y here represents the distance from the point (1, 1) to the original graph.

D. Use the calculator to find the minimum distance. Notice that this graph has two candidates for the minimum value. We are concerned with finding the absolute minimum distance. After some calculator work, we see that this occurs at the low point on the right side of the picture.

So we have that when x = 1.864 (approx.), the minimum distance d he point on the original graph which is

e inal equation . The resulting closest point is

approximately (1.864, 0.884). Check the original graph to see if the answer makes sense. ƒ

Here’s a more involved example for you. Multipart problems like these often appear in Calculus courses.

is approximately 0.872. T

closest to (1, 1) is now determined by plugging the x-value into th

3 _x

x y = −3 orig

Example 5 (Minimizing time)

A certa

second. He is located 20 feet away from the edge of the ocean. A swimmer is drowning 5

and the drowning victim is er the wate o

that he min er?

Solutio

in lifeguard can run at 15.7 feet per second and can swim at 5.9 feet per 0 feet out to sea. If the distance along the shore between the lifeguard

40 feet, where should the lifeguard ent r s imizes the time that it takes to reach the drowning swimm

n: S water 50 A x E B 20 feet 40 - x L land

his problem requires us to combine several of the ideas used earlier. To et up the problem let’s examine the picture and label everything. Let A be

e point on the shore closet to the lifeguard L, and let B be the point on the ho life B is 40 then 40 it will t rmula feet T s th

s re closest to the swimmer S. Let E be the point on the shore where the guard will enter the water. We are given that the distance between A and

feet. Let x be the distance from A to E. The distance from E to B is – x. We need to find a formula in terms of x for the amount of time ake for the lifeguard to reach the swimmer. We use the standard

RT

D= , in the form R D T =

fo , to find expressions for the time he

pends running and for the time that he spends swimming. The total time ill be the sum of these two expressions.

s w

A. Time spent running: His speed (rate) is 15.7. He runs an unknown distance d from point L to point E. The Pythagorean Theorem tells us that d2 =202 +x2, so that d = 202 +x2 . Then the time he spends running from L to E is 202 x2

r

d ₌ +

. Time spent swimming:

7 . 15

B. His speed is 5.9 and the distance from E to S is

(

)

50

40− x + . Then the time that he spends swimming

(

)

the total time, expressed in terms of x. C. Combine these to find

(

)

T . The question is: for whic

value is the time T at a minimum?

D. Graph 20

(

)

ulator and use the minimum finder to obtain x = 24.724771 and y = 10.886787 . So he should enter the water approximately 24.72 feet down shore to reach the swimmer in a minimum time of approximately 10.89 seconds. ƒ

quantity that you want to maxim or mini e. Th sho be written in terms of a single unkno

If the formula (equation) is quadratic then you can find the vertex algebraically. However, you can always approximate the max/min by using a graphing calculator. It is im different variables of the ordered pairs on the graph represent, and what you are asked to find. It is easy to get confused in these problem

calc

In all of these problems the general strategy is to first find a formula for the

ize miz is uld

s step is the most difficult aspect of these proble

wn quantity. Thi ms.

portant to know what the two s and to misinterpret the values.

3

a minimum.

Exercises 2.

Find the vertex of the following quadratic equations and state whether it is a maximum or 1. 2. 3. 4. 7 3 15 2 _{− +} = x x y 3 2 5 + = x x y ₋ 2 ₊ 01 . _x2 _{0.23 0.15} 0 − − − = x y 14 102 57 8 3 2 _{− +} = x x y

Find the m ound

your answers to 2 decimal places.) 5. a. b. 6. a. –15 < x < –3 Maximum b.

Solve the follo from Example

7. A ball 96-foot-high tower with an initial

velocity of 80 feet per second. When does the ball reach its maximum

, where x is the horizontal distance (in feet) from the foot of the embankment to a point directly under the shell. How high does the shell go, and how far away does it hit the water? (Hint: How high will it be when it hits the water?)

10. Based on data from past years, a consultant informs Tim’s Bicycles that their profit from selling x bicycles is given by the equation

aximum or minimum value for y given the restrictions on x. (R

0 < x < 3 Minimum –2 < x < 0 Maximum x x x y ₌ 3₋ 2 ₋6 450 145 − = x y _x3_{+ x}8 2 ₋ 0 < x < 10 Minimum

wing problems. For problems 7 and 8, use the height equation 1.

is thrown upward from the top of a height and how high is it at that time?

8. A ball is thrown upward from a height of 6 feet with an initial velocity of 32 feet per second. Find its maximum height.

9. During the Civil War, the standard heavy gun for coastal artillery was the 15-inch Rodman cannon, which fired a 330-pound shell. If one of these guns is fired from the top of a 50-foot-high shoreline embankment, then the height of the shell above the water (in feet) can be approximated by the equation h=−.0000167x2 +.23x+50 15000 4 250 2 − − = x x

p . How much profit do they make by selling 100 bicycles? How many bicycles should be sold to maximize the profit? 11. An auto parts manufacturer makes radiators that sell for $350 each. The

profit generated by selling x radiators is approximated by the equation . What number of radiators will produce the largest possible profit? What is the largest possible profit?

000 , 600 375 01 . 2 _{+ −} − = x x P

12. A farmer has 500 feet of fencing and wants to fence in a rectangular area traight river. What are the dimensions of the field with largest possible area, assuming that no fencing is needed along the river?

ting out equal size squares from the corners and folding up the sides. What size squares should be removed in order to produce a box with maximum volume?

An open-top box with a square base is to be constructed from 150 sq cm of material. What dimensions for the box will produce the largest possible volume? (Hint: You need to find a formula for the height of the box given that the surface area is 150 sq cm.)

m the graph of to the point (1, 2).

17. For in the first quadrant and is on the graph 2_{, consider the rectangle with corners (–x, 0), (–x, y), (x, 0), and}

this rectangle have maximal area?

h

ach pter. Where should she leav

next to a s

13. An open-top rectangular box is to be made from a 35-inch by 35-inch piece of sheet metal by cut

14.

15. Find the minimum distance from the graph of y= x2 to the point (3, 0). 16. Find the minimum distance fro y =x3 +x2 +6

every point (x, y) which is 4

24 x

y = −

(x, y). For which value of x does

helicopter

Anne 1 mile

2 miles

18. Anne is standing on a straight road and wants to reach her helicopter, whic is located 2 miles down road from her, then a mile to the right in a field. She can run 5 miles per hour on the road and 3 miles per hour in the field. She plans to run down the road, then cut diagonally across the field to re the helico e the road in order to reach the helicopter in exactly 42 minutes? What is the least amount of time that it will take her to reach the helicopter? (Hint: See example 4.)