   Chapter 10.4, Problem 19E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Optimization at a fixed cost A rectangular area is to be enclosed and divided into thirds. The family has $800 to spend for the fencing material. The outside fence costs$10 per foot installed, and the dividers cost $20 per foot installed. What are the dimensions that will maximize the area enclosed? (The answer contains a fraction.) To determine To calculate: The dimensions to maximize the area if the rectangular area is divided into three parts. Explanation Given Information: The rectangular area is divided into three parts such that the total cost of fencing is$800 ($10 per foot for the outside boundary and$20 per foot for the dividers).

Formula used:

To find the maximum value, calculate the relevant stationary value of an equation. Differentiate the function with respect to the independent variable and equate it to 0. the relevant value is the stationary value that satisfies the provided conditions.

If the second derivative of the provided function is less than zero, then substituting the value of the independent variable will give the maximum value of the equation.

The power rule is used for a function in which the expression can be written as every term raised to a power (be it fractional, positive or negative). For the function f(x)=xn, the derivative is

ddx[xn]=nxn1

Calculation:

Consider the provided statement,

The rectangular area is divided into three parts such that the total cost of fencing is $800 for which the cost$10 per foot for the outside boundary and \$20 per foot for the dividers.

In the equation A=lb, there are two independent variables l and b. Express the equation as a function of one variable by using the provided conditions.

Assume P is the perimeter of the rectangular field.

To fence the fields, the perimeter P=2(l+b)+2b as partition requires two fences.

Use the equation for cost for fencing is 800=10(2l+2b)+20(2b) and make b the subject to get

800=20l+60bb=80020l60

Apply the equation A=lb and substitute b=80020l60 l to get

A=l[

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