The complete area of a page on the book is 90 in^2. The area that can be printed on is less than this by the area of the margins. The page has 1 inch margins at the top and bottom and 1/2 inch margins at the sides.

Let the length...

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The complete area of a page on the book is 90 in^2. The area that can be printed on is less than this by the area of the margins. The page has 1 inch margins at the top and bottom and 1/2 inch margins at the sides.

Let the length of the page be L. Length is the dimension of the side. As the area of the page is 90, the width is 90/L. Width is the dimension of the top and bottom. The side margins have an area 2*(1/2)*L. The top and bottom margins have an area (90/L - 1)*2*1

The total area occupied by the margins is 2*(1/2)*L + (90/L - 1)*2*1

=> L + 180/L - 2

This has to be minimized. Equating the derivative to 0 and solving gives:

[L + 180/L - 2]' = 0

=> 1 - 180/L^2 = 0

=> 180 = L^2

=> L = `6*sqrt5`

The width is `90/(6*sqrt 5) = 3*sqrt5`

**The required dimensions of the page are a length of `6*sqrt 5 ` and a width of `3*sqrt 5` **