Question

A rectangular page is designed to contain 81 square inches of print. The margins at the top and bottom of the page are each 1 inch deep. The margins on each side are 1½ inches wide. What should the dimensions of the page be so that the least amount of paper is used?

Answer 1

Answer:

x = 11.02 inch by  y = 7.35 inch

Explanation:

Let the width of the print be = x

Then the width of the page is x + (1.5+1.5) = x + 3

Let the height of the print be = y

Then the height of the page is y + (1+1) = y + 2

Given the area of the print is 81 square inches

That means, xy = 81

Therefore, y = $\frac{81}{x}$

Therefore, the area of the page is

A = (x + 3)(y + 2)

   = (x + 3)($\frac{81}{x}$ + 2)

Now for least amount of paper to be used,

$\frac{dA}{dx}=(x+3)(\frac{81}{x}+2)$

$\frac{dA}{dx}= \left (\frac{81}{x}+2 \right )+\left ( x+3 \right )\left ( -\frac{81}{x^{2}} \right )=0$

$\frac{81+2x}{x}=\frac{(x+3)(81)}{x^{2}}$

$81x+2x^{2}=81x+81\times 3$

$2x^{2}=243$

x = 11.02 inch

Now y = $\frac{81}{x}$

          = $\frac{81}{11.02}$

          = 7.35 inch

Therefore the dimensions of thr page for the least amount of paper is x = 11.02 inch by  y = 7.35 inch