   Chapter 3.P, Problem 2P

Chapter
Section
Textbook Problem

# Show that x 2 y 2 ( 4 − x 2 ) ( 4 − y ) 2 ≤ 16 for all numbers x and y such that | x | ≤ 2 and | y | ≤ 2 .

To determine

To show:

x2y24-x24-y216 for all x and y such that x2 and y2.

Explanation

1) Concept:

Use Closed Interval Method to find the absolute maximum and absolute minimum values for the given function.

2) The Closed Interval Method:

To find the absolute maximum and minimum values of a continuous function f on a closed interval a, b:

i. Find the values of f at the critical numbers of f in a, b.

ii. Find the values of f at the endpoints of the interval.

iii. The largest of the values from the above steps is the absolute maximum value and the smallest of these values is the absolute minimum value.

3) Calculation:

To show that

x2y24-x24-y216

Let fxy=x2y24-x24-y2

Take like terms together.

fxy=x24-x2y24-y2

Now let X=x24-x2 and Y=y24-y2

Then above function becomes

fXY=XY

Take derivative of X=x24-x2.

X'=x2-2x+4-x2(2x)

Simplify.

X'=-2x3+8x-2x3

Combine like terms

X'=-4x3+8x

To find critical numbers, equate X'=0,

-4x3+8x=0

Simplify.

-4xx2-2=0

x=0 or x2=2

x=0 or x=±2

Therefore, critical numbers are 0, ±2 .

To find the absolute maximum and minimum values,

Substitute critical numbers and endpoints in the function f.

Put x=0and ±2  in fx=x24-x2

f0=0

f±2=±224-(±2)2

Simplify

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