/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 A rectangular page is to contain... [FREE SOLUTION] | 91Ó°ÊÓ

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A rectangular page is to contain 30 square inches of print. The margins at the top and bottom of the page are to be 2 inches wide. The margins on each side are to be 1 inch wide. Find the dimensions of the page such that the least amount of paper is used.

Short Answer

Expert verified
The page's dimensions that would minimize the use of paper are the width found in step 3 and the length found in step 5.

Step by step solution

01

Define Variables

Denote the width of the page as \( w \), and the length of the page as \( l \). The print area is \( (w-2) \times (l-4) = 30 \) square inches, which gives expression of length using the width, \( l = 4 + \frac{30}{w-2} \).
02

Obtain Area

As the objective is to minimize the area of the page, formulate the area of the page function \( A(w) = w \times l \) using the expression found for the length. Therefore, \( A(w) = w \times (4 + \frac{30}{ w-2}) = 4w + \frac{30w}{w-2} \).
03

Minimize Area

To minimize the area, differentiate the area function w.r.t width and equate it to zero to find the critical points. The derivative \( A'(w) = 4 + \frac{30}{w-2} - \frac{60}{(w-2)^2} \). To solve \( A'(w) = 0 \) for \( w \) yields the width of the page.
04

Test Critical Points

Substitute derivative's critical points and the endpoints ( \( w = 2 \) will be the smallest possible width allowed due to margins) into second derivative \( A''(w)=\frac{120}{(w-2)^3} \) to determine the minimum. A positive value means the function is concave up there and we have found a minimum.
05

Find Length

Now that we know the width that minimizes the area, substitute \( w \) into \( l = 4 + \frac{30}{ w-2} \) to find the corresponding length.

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