# To prove : For all values of x and y such that, | x | ≤ 2 and | y | ≤ 2 , x 2 y 2 ( 4 − x 2 ) ( 4 − y 2 ) ≤ 16 .

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 4, Problem 4P
To determine

## To prove: For all values of x and y such that, |x|≤2 and |y|≤2, x2y2(4−x2)(4−y2)≤16.

Expert Solution

### Explanation of Solution

Proof:

Consider the functions, f(t)=t2(4t2).

Then, the given function can be written as, x2y2(4x2)(4y2)=f(x)f(y)

Find absolute minimum and absolute maximum of the function f(t)=t2(4t2) over the interval [2,2].

Differentiate f(t) with respect to t.

f(t)=8t4t3.

Set f(t)=0 and obtain the critical points over the interval [2,2].

8t4t3=04t(24t2)=04t=0,t2=2t=0,±2

Thus, the critical points are 2,0and 2.

Apply the extreme values of the given interval and the critical number in f(t).

Substitute t=2 in f(t),

f(2)=(2)2(4(2)2)=4(0)=0

Substitute t=2 in f(t),

f(2)=(2)2(4(2)2)=2(42)=4

Substitute t=0 in f(t),

f(0)=02(402)=0(4)=0

Substitute t=2 in f(t),

f(2)=(2)2(4(2)2)=2(42)=4

Substitute t=2 in f(t),

f(2)=22(422)=4(0)=0

Since the largest functional value is the absolute maximum and the smallest functional value is the absolute minimum, the absolute maximum of f(t) is 4 and the absolute minimum of f(t) is 0.

Hence, over the interval [2,2], 0f(t)4.

Similarly, 0f(x)4 and 0f(y)4.

Thus, the product of f(x) and f(y) is lies between,

0f(x)f(y)160x2y2(4x2)(4y2)16

Therefore, it is proved that for |x|2 and |y|2, 0x2y2(4x2)(4y2)16.

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