# The table whose perimeter is 2400 and area of each configuration is maximum. ### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071 ### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 2, Problem 21P

a.

To determine

## The table whose perimeter is 2400 and area of each configuration is maximum.

Expert Solution

The possible pairs whose perimeter is 2400 and area is maximum is length =1200 and breadth=600.

### Explanation of Solution

Given information:

2400ft of fencing, that should be fenced off a rectangular field.

Concept Used:

Let x is the breadth of rectangular field and y is the length of rectangular field.

Then 2x+y=2400     (by fencing three sides)

y is the side parallel to river and x is the other side.

Calculation:

 X Y Product 100 2200 210000 200 2000 40000 300 1800 540000 400 1600 640000 500 1400 700000 600 1200 720000 700 1000 700000

Conclusion:

Only these values are possible after that area start decreasing. So answer for maximum area 600 is breadth and 1200 is length.

b.

To determine

### To determine the area in terms of one side.

Expert Solution

The area is 24002x2

### Explanation of Solution

Given information :

Perimeter is 2400ft and area is maximum.

Concept Used:

Perimeter is 2400ft and area is maximum.

Let length of rectangular field be y and breadth be x.

Calculation:

Fencing along three sides of river.

2x+y=2400ftxy=area12x+y=2400y=24002x2

Put equation 2 in 1,

area=x(24002x)area=24002x2

Conclusion:

The required area in terms of one side is 24002x2

c.

To determine

### To compare the answer with subpart a .

Expert Solution

The answer is same as subpart a .

### Explanation of Solution

Given information :

Perimeter is 2400ft and area is maximum

Concept used:

Maximise the area by using differentiation.

First differentiation is equal to zero for maximization.

Calculation:

area=24002x2A(x)=2400x2x2

Differentiate with respect to x ,

A'(x)=24004x

Put A'(x)=0 for maximum value

24004x=0x=600Then y=1200    (Using equation 1)

Conclusion:

The answer is same as subpart a . So model of function is correct.

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