04 Lagrange Multiplier.pdf

Lagrange Multiplier

Lucky Galvez

Institute of Mathematics University of the Philippines

Recall

R (a, b) is a critical point off(x, y) if ∇f(a, b) =~0

(fx(a, b) =fy(a, b) = 0) or either one offx(a, b) or fy(a, b)

is undefined.

R For a critical point (a, b) off, define

D=fxx(a, b)fyy(a, b)−[fxy(a, b)]2. Then f has a relative minimum at (a, b) ifD >0 and

fxx(a, b)>0.

f has a relative maximum at (a, b) ifD >0 and

fxx(a, b)<0.

f has a saddle point at (a, b) ifD <0.

R If f is defined on a closed and bounded regionR⊂_R2_, then f attains its abolute extrema at the critical points in

R or at the boundary of R

Lagrange Multiplier

Suppose we want to find the maximum value off(x, y) over all points (x, y) satisfying g(x, y) =k (e.g., we want to maximize volume of a box such that its surface area is a constantk).

Consider the level curvesf(x, y) =c, for different values ofc

and think ofg(x, y) =kas a particular level curve for a fixedk. We are actually looking for the largestcsuch that f(x, y) =c

intersectsg(x, y) =k.

f(x, y) =c6

f(x, y) =c1

f(x, y) =c5

f(x, y) =c2

f(x, y) =c4

f(x, y) =c3

g(x, y) =k

Intuitively, this happens when they are tangent to each other, i.e., ∃λ such that

∇f(x, y) =λ∇g(x, y)

Lagrange Multiplier

Suppose we want to find the maximum value off(x, y) over all points (x, y) satisfying g(x, y) =k (e.g., we want to maximize volume of a box such that its surface area is a constantk). Consider the level curvesf(x, y) =c, for different values ofc

and think ofg(x, y) =kas a particular level curve for a fixedk. We are actually looking for the largestcsuch that f(x, y) =c

intersectsg(x, y) =k.

f(x, y) =c6

f(x, y) =c1

f(x, y) =c5

f(x, y) =c2

f(x, y) =c4

f(x, y) =c3

g(x, y) =k

Intuitively, this happens when they are tangent to each other, i.e., ∃λ such that

∇f(x, y) =λ∇g(x, y)

Suchλis called theLagrange multiplier.

Lagrange Multiplier

Suppose we want to find the maximum value off(x, y) over all points (x, y) satisfying g(x, y) =k (e.g., we want to maximize volume of a box such that its surface area is a constantk). Consider the level curvesf(x, y) =c, for different values ofc

and think ofg(x, y) =kas a particular level curve for a fixedk. We are actually looking for the largestcsuch that f(x, y) =c

intersectsg(x, y) =k.

f(x, y) =c6

f(x, y) =c1

f(x, y) =c5

f(x, y) =c2

f(x, y) =c4

f(x, y) =c3

Intuitively, this happens when they are tangent to each other, i.e., ∃λ such that

∇f(x, y) =λ∇g(x, y)

Lagrange Multiplier

Suppose we want to find the maximum value off(x, y) over all points (x, y) satisfying g(x, y) =k (e.g., we want to maximize volume of a box such that its surface area is a constantk). Consider the level curvesf(x, y) =c, for different values ofc

and think ofg(x, y) =kas a particular level curve for a fixedk. We are actually looking for the largestcsuch that f(x, y) =c

intersectsg(x, y) =k.

f(x, y) =c6

f(x, y) =c1

f(x, y) =c5

f(x, y) =c2

f(x, y) =c4

f(x, y) =c3

g(x, y) =k

Intuitively, this happens when they are tangent to each other

, i.e., ∃λ such that

∇f(x, y) =λ∇g(x, y)

Suchλis called theLagrange multiplier.

Lagrange Multiplier

Suppose we want to find the maximum value off(x, y) over all points (x, y) satisfying g(x, y) =k (e.g., we want to maximize volume of a box such that its surface area is a constantk). Consider the level curvesf(x, y) =c, for different values ofc

and think ofg(x, y) =kas a particular level curve for a fixedk. We are actually looking for the largestcsuch that f(x, y) =c

intersectsg(x, y) =k.

f(x, y) =c6

f(x, y) =c1

f(x, y) =c5

f(x, y) =c2

f(x, y) =c4

f(x, y) =c3

Intuitively, this happens when they are tangent to each other, i.e., ∃λ such that

∇f(x, y) =λ∇g(x, y)

Lagrange Multiplier

Suppose we want to find the maximum value off(x, y) over all points (x, y) satisfying g(x, y) =k (e.g., we want to maximize volume of a box such that its surface area is a constantk). Consider the level curvesf(x, y) =c, for different values ofc

and think ofg(x, y) =kas a particular level curve for a fixedk. We are actually looking for the largestcsuch that f(x, y) =c

intersectsg(x, y) =k.

f(x, y) =c6

f(x, y) =c1

f(x, y) =c5

f(x, y) =c2

f(x, y) =c4

f(x, y) =c3

g(x, y) =k

Intuitively, this happens when they are tangent to each other, i.e., ∃λ such that

∇f(x, y) =λ∇g(x, y)

Suchλis called theLagrange multiplier.

Lagrange Multiplier

Method of Lagrange Multiplier

Suppose the extreme values off(x, y) exist and∇g6= 0 on the surfaceg(x, y) =k. To find the extreme values off(x, y) subject to the constraintg(x, y) =k,

1 Find all solutions to the system (

∇f(x, y) =λ∇g(x, y)

g(x, y) =k .

2 _{Evaluate f at each of the solutions. The largest(smallest)}

value is the absolute maximum(minimum) value of f

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

(maximum) f(−10,6) =−68 (minimum)

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k

⇔ 





5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

(maximum) f(−10,6) =−68 (minimum)

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

(maximum) f(−10,6) =−68 (minimum)

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5 2λ

and from (2): y=− 3

2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

(maximum) f(−10,6) =−68 (minimum)

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136

⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136

⇒ 1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

(maximum) f(−10,6) =−68 (minimum)

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16

⇒λ=±1

4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

(maximum) f(−10,6) =−68 (minimum)

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x=

10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y=

−6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

(maximum) f(−10,6) =−68 (minimum)

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6.

Ifλ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1 4, x=

−10,y= 6. Hence,

f(10,−6) = 68

(maximum) f(−10,6) =−68 (minimum)

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y=

6. Hence,

f(10,−6) = 68

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

(maximum) f(−10,6) =−68 (minimum)

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68

(maximum)

f(−10,6) =−68

(minimum)

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

Example

Example

Find the extreme values off(x, y) = 5x−3y subject to the constraintx2_+y2_{= 136.}

Solution. Letf(x, y) = 5x−3y andg(x, y) =x2_+y2_{. We solve}

the system

(

∇f(x, y) =λ∇g(x, y)

g(x, y) =k ⇔







5 = 2λx (1)

−3 = 2λy (2)

x2_+y2 _{= 136 (3)}

From (1): x= 5

2λ and from (2): y=−

3 2λ.

Substituting in (3):

₅

2λ 2

+

− 3

2λ 2

= 136⇒ 34

4λ2 = 136⇒

1

λ2 = 16⇒λ=±

1 4

Ifλ= 1₄,x= 10,y= −6. If λ=−1

4, x= −10,y= 6.

Hence,

f(10,−6) = 68 (maximum) f(−10,6) =−68 (minimum)

Exercises

1 Find the extreme values of the function subject to the

given constraint.

a. g(x, y) =e−xy; x2_{+ 4y}2_{= 1}

b. h(x, y, z) =x2_y2_z2_;x2_+y2_+z2_{= 1}

2 _{Find the extreme values of f(x, y) = 2x}2_{+ 3y}2−_4x−_{5 on}

the closed disk centered at the origin with radius 4.

3 Find the lowest point on the curve of intersection of the

plane 4y−x= 5 and the paraboloidz=x2+y2.

4 Find the dimensions of the rectangular box without a lid

having the largest volume if the total surface area is 12cm2.

5 _{Find the maximum volume of a rectangular box inscribed}

in a sphere of radius r.

6 Use the Method of Lagrange multipliers to prove that

Exercises

1 Find the extreme values of the function subject to the

given constraint.

a. g(x, y) =e−xy; x2_{+ 4y}2_{= 1}

b. h(x, y, z) =x2_y2_z2_;x2_+y2_+z2_{= 1}

2 _{Find the extreme values of f(x, y) = 2x}2_{+ 3y}2−_4x−_{5 on}

the closed disk centered at the origin with radius 4.

3 Find the lowest point on the curve of intersection of the

plane 4y−x= 5 and the paraboloidz=x2+y2.

4 Find the dimensions of the rectangular box without a lid

having the largest volume if the total surface area is 12cm2.

5 _{Find the maximum volume of a rectangular box inscribed}

in a sphere of radius r.

6 Use the Method of Lagrange multipliers to prove that

rectangle the with the maximum area having a given perimeterp is a square. Find the length of a side in terms of p.

Exercises

1 Find the extreme values of the function subject to the

given constraint.

a. g(x, y) =e−xy; x2_{+ 4y}2_{= 1}

b. h(x, y, z) =x2_y2_z2_;x2_+y2_+z2_{= 1}

2 _{Find the extreme values of f(x, y) = 2x}2_{+ 3y}2−_4x−_{5 on}

the closed disk centered at the origin with radius 4.

3 Find the lowest point on the curve of intersection of the

plane 4y−x= 5 and the paraboloidz=x2+y2.

4 Find the dimensions of the rectangular box without a lid

having the largest volume if the total surface area is 12cm2.

5 _{Find the maximum volume of a rectangular box inscribed}

in a sphere of radius r.

6 Use the Method of Lagrange multipliers to prove that

Exercises

1 Find the extreme values of the function subject to the

given constraint.

a. g(x, y) =e−xy; x2_{+ 4y}2_{= 1}

b. h(x, y, z) =x2_y2_z2_;x2_+y2_+z2_{= 1}

2 _{Find the extreme values of f(x, y) = 2x}2_{+ 3y}2−_4x−_{5 on}

the closed disk centered at the origin with radius 4.

3 Find the lowest point on the curve of intersection of the

plane 4y−x= 5 and the paraboloidz=x2+y2.

4 Find the dimensions of the rectangular box without a lid

having the largest volume if the total surface area is 12cm2.

5 _{Find the maximum volume of a rectangular box inscribed}

in a sphere of radius r.

6 Use the Method of Lagrange multipliers to prove that

rectangle the with the maximum area having a given perimeterp is a square. Find the length of a side in terms of p.

Exercises

1 Find the extreme values of the function subject to the

given constraint.

a. g(x, y) =e−xy; x2_{+ 4y}2_{= 1}

b. h(x, y, z) =x2_y2_z2_;x2_+y2_+z2_{= 1}

2 _{Find the extreme values of f(x, y) = 2x}2_{+ 3y}2−_4x−_{5 on}

the closed disk centered at the origin with radius 4.

3 Find the lowest point on the curve of intersection of the

plane 4y−x= 5 and the paraboloidz=x2+y2.

4 Find the dimensions of the rectangular box without a lid

having the largest volume if the total surface area is 12cm2.

5 _{Find the maximum volume of a rectangular box inscribed}

in a sphere of radius r.

6 Use the Method of Lagrange multipliers to prove that

Exercises

1 Find the extreme values of the function subject to the

given constraint.

a. g(x, y) =e−xy; x2_{+ 4y}2_{= 1}

b. h(x, y, z) =x2_y2_z2_;x2_+y2_+z2_{= 1}

2 _{Find the extreme values of f(x, y) = 2x}2_{+ 3y}2−_4x−_{5 on}

the closed disk centered at the origin with radius 4.

3 Find the lowest point on the curve of intersection of the

plane 4y−x= 5 and the paraboloidz=x2+y2.

4 Find the dimensions of the rectangular box without a lid

having the largest volume if the total surface area is 12cm2.

5 _{Find the maximum volume of a rectangular box inscribed}

in a sphere of radius r.

6 Use the Method of Lagrange multipliers to prove that

rectangle the with the maximum area having a given perimeterp is a square. Find the length of a side in terms of p.

Exercises

1 Find the extreme values of the function subject to the

given constraint.

a. g(x, y) =e−xy; x2_{+ 4y}2_{= 1}

b. h(x, y, z) =x2_y2_z2_;x2_+y2_+z2_{= 1}

2 _{Find the extreme values of f(x, y) = 2x}2_{+ 3y}2−_4x−_{5 on}

the closed disk centered at the origin with radius 4.

3 Find the lowest point on the curve of intersection of the

plane 4y−x= 5 and the paraboloidz=x2+y2.

4 Find the dimensions of the rectangular box without a lid

having the largest volume if the total surface area is 12cm2.

5 _{Find the maximum volume of a rectangular box inscribed}

in a sphere of radius r.

6 Use the Method of Lagrange multipliers to prove that

References

1 _{Stewart, J., Calculus, Early Transcendentals, 6 ed., Thomson}

Brooks/Cole, 2008

2 _{Leithold, L.,The Calculus 7, Harper Collins College Div., 1995}

3 _{Dawkins, P.,Calculus 3, online notes available at}

http://tutorial.math.lamar.edu/