   Chapter 10.4, Problem 17E ### 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

# Minimum cost A rectangular field with one side along a river is to be fenced. Suppose that no fence is needed along the river, the fence on the side opposite the river costs $20 per foot, and the fence on the other sides costs$5 per foot. If the field must contain 45,000 square feet, what dimensions will minimize costs?

To determine

To calculate: The dimensions of rectangular field to minimize the costs.

Explanation

Given Information:

The field contains 45,000 sq.ft with fencing required on 3 sides (two sides at $5 per foot and third side at$20 per foot).

Formula used:

To find the minimum 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 more than zero, then substituting the value of the independent variable will give the minimum 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:

As given, the area of the rectangular field is 45,000 sq.ft with fencing required on 3 sides

The fencing of two sides at $5 per foot and third side at$20 per foot.

Assume, the perimeter of the rectangular field is P

So, the perimeter is P=l+2b.

The cost for fencing is C=20l+5(2b).

In the equation C=20l+5(2b), there are two independent variables l and b. Express the equation as a function of one variable by using the provided conditions.

Use the equation for area lb=45,000 and make b the subject to get

b=45000l

Apply the equation C=20l+5(2b) and substitute b=45000l to get

C=20l+5

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