   Chapter 14.7, Problem 34E

Chapter
Section
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

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Evaluate the definite integral. 011+7x3dx

Single Variable Calculus: Early Transcendentals, Volume I

What are the two requirements for a random sample?

Statistics for The Behavioral Sciences (MindTap Course List)

In Problems 39 – 44, solve each inequality and graph the solution. 40.

Mathematical Applications for the Management, Life, and Social Sciences

The radius of convergence of is: 1 3 ∞

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th

Write in cylindrical coordinates, where E is the solid at the right.

Study Guide for Stewart's Multivariable Calculus, 8th 