Chapter 10.4, Problem 21E

### Mathematical Applications for the ...

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

Chapter
Section

### Mathematical Applications for the ...

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

# Minimum cost The base of a rectangular box is to be twice as long as it is wide. The volume of the box is 256 cubic inches. The material for the top costs $0.10 per square inch and the material for the sides and bottom costs$0.05 per square inch. Find the dimensions that will make the cost a minimum.

To determine

To calculate: The dimensions to minimize the cost of the box.

Explanation

Given Information:

The volume of the box is 256 cubic inches and base is twice as long as the width. The cost of the material at the top is $0.50 per sq. inch and for the sides/bottom is 0.25 per square inch. 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. Check for local minima, 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 Volume of a rectangular box is given by: V=length×breadth×height Calculation: Consider the provided statement, The volume of the box is 256 cubic inches and base is twice as long as the width. The cost of the material at the top is$0.50 per sq. inch and for the sides/bottom is 0.25 per square inch.

Assume l is length, b is breadth or base and his height

Thus, volume of the box is V=l×b×h

When base is twice as long as the width, the volume 256=l(2l)h.

The cost for fencing is C=0.50l(2l)+0.25[l(2l)+h(2l)(2)+lh(2)].

In the equation C=0.50l(2l)+0.25[l(2l)+h(2l)(2)+lh(2)], 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 256=l(2l)h and make h the subject to get

h=2562l2h=128l2

Apply the equation C=0.50l(2l)+0.25[l(2l)+h(2l)(2)+lh(2)] and substitute h in terms of l to get

C=l2+0.25[2l2+512l+256l]C=1

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