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Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643

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Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643
Textbook Problem

Find the absolute maximum and minimum values of f on the set D.

35. f(x, y) = x2 + 2y2 − 2x – 4y + 1, D = {(x, y) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 3}

To determine

To find: The absolute maximum and minimum values of the function f(x,y)=x2+2y22x4y+1 on the disk, D={(x,y)|0x2,0y3} .

Explanation

Given:

The function is f(x,y)=x2+2y22x4y+1 on the disk D={(x,y)|0x2,0y3} .

Calculation:

Take the partial derivative in the given function with respect to x and obtain fx .

fx=x(x2+2y22x4y+1)=x(x2)+x(2y2)x(2x)x(4y)+x(1)=2x+02(1)0+0=2x2

Thus, fx=2x2 . (1)

Take the partial derivative in the given function with respect to y and obtain fy .

fy=y(x2+2y22x4y+1)=y(x2)+y(2y2)y(2x)y(4y)+y(1)=0+2(2y)04(1)+0=4y4

Thus, fy=4y4 . (2)

Set the above partial derivatives to 0 and find the values of x,y.

From the equation (1),

2x2=02x=2x=1

From the equation (2),

4y4=04y=4y=1

Thus, the critical point is (1,1) .

Substitute x=1,y=1 in the function f(x,y) .

f(1,1)=(1)2+2(1)22(1)4(1)+1=1+224+1=24=2

Thus, f(1,1)=2 .

The disk D={(x,y)|0x2,0y3} is shown below in Figure 1.

Using two point formula, the equation of L1 is y=0 , 0x2 .

The equation of L2 is x=2 , 0y3

The equation of L3 is y=3 , 0x2 .

The equation of L4 is x=0 , 0y3 .

Along the L1 , y=0 .

Substitute y=0 in the function f(x,y) ,

f(x,0)=x2+2(0)22x4(0)+1=x2+02x0+1=x22x+1

Differentiate the above equation.

f'(x,0)=2x2(1)+0=2x2

Equate the above derivative to 0.

2x2=02x=2x=22x=1

Thus, the critical point along the line L1 is (1,0) and the endpoints are (0,0),(2,0) .

Substitute the endpoints and the critical point of the line L1 in the function f(x,y) and find its functional values.

Substitute x=1,y=0 in the function f(x,y) .

f(1,0)=(1)2+2(0)22(1)4(0)+1=1+020+1=22=0

Hence, f(1,0)=0

Substitute x=0,y=0 in the function f(x,y) .

f(0,0)=(0)2+2(0)22(0)4(0)+1=1

Thus, f(0,0)=1 .

Substitute x=2,y=0 in the function f(x,y) .

f(2,0)=(2)2+2(0)22(2)4(0)+1=4+040+1=1

Hence, f(2,0)=1 .

Along the line L2 , x=2 .

Substitute x=2 in the function f(x,y) ,

f(2,y)=(2)2+2y22(2)4y+1=4+2y244y+1=2y24y+1

Differentiate the above equation,

f'(2,y)=2(2y)4(1)+0=4y4

Equate the above derivative to 0

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Chapter 14 Solutions

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