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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.

34. f(x, y) = x2 + xy + y2 − 6y, D = {(x, y) | −3 ≤ x ≤ 3, 0 ≤ y ≤ 5}

To determine

To find: The absolute maximum and minimum values of the function f(x,y)=x2+xy+y26y on the disk, D={(x,y)|3x3,0y5} .

Explanation

Given:

The function is f(x,y)=x2+xy+y26y on the disk, D={(x,y)|3x3,0y5} .

Calculation:

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

fx=x(x2+xy+y26y)=x(x2)+x(xy)+x(y2)x(6y)=2x+y(1)+00=2x+y

Thus, fx=2x+y . (1)

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

fy=y(x2+xy+y26y)=y(x2)+y(xy)+y(y2)y(6y)=0+x(1)+2y6(1)=x+2y6

Thus, fy=x+2y6 . (2)

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

From the equation (1),

2x+y=0y=2x

Substitute y=2x in the equation (2),

x+2(2x)6=0x4x63x=6x=2

Substitute x=2 in the equation y=2x ,

y=2(2)=4

Thus, the critical point is (2,4) .

Substitute x=2,y=4 in the function f(x,y) .

f(2,4)=(2)2+(2)(4)+(4)26(4)=48+1624=2032=12

Hence, f(2,4)=12 .

The disk D={(x,y)|3x3,0y5} is shown below in Figure 1.

Using two point formula, the equation of L1 is y=0 , 3x3 .

The equation of L2 is x=3 , 0y5

The equation of L3 is y=5 , 3x3 .

The equation of L4 is x=3 .

Along the line L1 , the value of y=0 .

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

f(x,0)=x2+x(0)+(0)26(0)=x2+0+00=x2

Differentiate the above equation, f'(x,0)=2x .

Equate the above derivative to 0.

2x=0x=0

Thus, the critical point along the line L1 is (0,0) and it endpoints are (3,0),(3,0) .

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

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

f(0,0)=(0)2+(0)(0)+(0)26(0)=0

Hence, f(0,0)=0 .

Substitute x=3,y=0 in the function f(x,y) .

f(3,0)=(3)2+(3)(0)+(0)26(0)=9+0+00=9

Thus, f(3,0)=9 .

Substitute x=3,y=0 in the function f(x,y) .

f(3,0)=(3)2+(3)(0)+(0)26(0)=9+0+00=9

Hence, f(3,0)=9 .

Along the L2 , the value of x=3 .

Substitute x=3 in the function f(x,y) ,

f(3,y)=(3)2+(3)y+y26y=9+3y+y26y=y23y+9

Differentiate the above equation.

f'(3,y)=2y3(1)+0=2y3

Equate the above derivative to 0

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Chapter 14 Solutions

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P-14RECh-14 P-15RECh-14 P-16RECh-14 P-17RECh-14 P-18RECh-14 P-19RECh-14 P-20RECh-14 P-21RECh-14 P-22RECh-14 P-23RECh-14 P-24RECh-14 P-25RECh-14 P-26RECh-14 P-27RECh-14 P-28RECh-14 P-29RECh-14 P-30RECh-14 P-31RECh-14 P-32RECh-14 P-33RECh-14 P-34RECh-14 P-35RECh-14 P-36RECh-14 P-37RECh-14 P-38RECh-14 P-39RECh-14 P-40RECh-14 P-41RECh-14 P-42RECh-14 P-43RECh-14 P-44RECh-14 P-45RECh-14 P-46RECh-14 P-47RECh-14 P-48RECh-14 P-49RECh-14 P-50RECh-14 P-51RECh-14 P-52RECh-14 P-53RECh-14 P-54RECh-14 P-55RECh-14 P-56RECh-14 P-57RECh-14 P-58RECh-14 P-59RECh-14 P-60RECh-14 P-61RECh-14 P-62RECh-14 P-63RECh-14 P-64RECh-14 P-65RECh-14 P-1PCh-14 P-2PCh-14 P-3PCh-14 P-4PCh-14 P-5PCh-14 P-6PCh-14 P-7PCh-14 P-8P

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