   Chapter 14, Problem 55RE

Chapter
Section
Textbook Problem

Find the absolute maximum and minimum values of f on the set D.55. f(x, y) = 4xy2 – x2y2 – xy3; D is the closed triangular region in the .xy-plane with vertices (0, 0), (0, 6), and (6, 0)

To determine

To find: The absolute maximum and minimum values of the function f(x,y)=4xy2x2y2xy3 on the closed triangular region D in the xy-plane with vertices (0,0),(0,6) and (6,0) .

Explanation

Given:

The function is f(x,y)=4xy2x2y2xy3 on the closed triangular region D with vertices (0,0),(0,6) and (6,0) .

Calculation:

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

fx=x(4xy2x2y2xy3)=x(4xy2)x(x2y2)x(xy3)=4y2(1)y2(2x)y3(1)=4y22xy2y3

Thus, fx=4y22xy2y3 (1)

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

fy=y(4xy2x2y2xy3)=y(4xy2)y(x2y2)y(xy3)=4x(2y)x2(2y)x(3y2)=8xy2x2y3xy2

Thus, fy=8xy2x2y3xy2 (2)

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

From the equation (1),

4y22xy2y3=0y2(42xy)=0y2=0,42xy=0y=0,y=42x

The value of y=0 is not inside the triangular region.

Substitute y=42x in the equation (2),

8x(42x)2x2(42x)3x(42x)2=032x16x28x2+4x33x(16+4x216x)=032x24x2+4x348x12x3+48x2=08x3+24x216x=0

Simplify further as follows.

8x(x23x+2)=08x=0,x23x+2=0x=0,(x2)(x1)=0x=0,x=2,x=1

Substitute x=0 in y=42x ,

y=42(0)=42=2

The point (0,2) is not inside the region.

Substitute x=2 in y=42x ,

y=42(2)=44=0

The point (2,0) is not inside the region.

Substitute x=1 in y=42x ,

y=42(1)=42=2

Thus,the critical point is (1,2) .

Substitute x=1,y=2 in the function f(x,y) .

f(1,2)=4(1)(2)2(1)2(2)2(1)(2)3=1648=1612=4

Hence, f(1,2)=4 .

Closed triangular region D, x y- plane with vertices (0,0),(0,6) and (6,0) is shown below in Figure 1.

Using two point formula, the equation of L1 is y=0 , 0x6 .

The equation of L2 is, x=y+6 .

The equation of L3 is, x=0 , 0y6 .

Along the line L1 , the value of y=0

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