BuyFind

Precalculus: Mathematics for Calcu...

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

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

Answer to Problem 21P

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
    1002200210000
    200200040000
    3001800540000
    4001600640000
    5001400700000
    6001200720000
    7001000700000

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

Answer to Problem 21P

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

Answer to Problem 21P

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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