# Math Help - Word Problem

1. ## Word Problem

A rectangular page is designed to contain 72 square inches of print. The margins at the top and bottom of the page are each 4 inches deep. The margins on each side are 2 inches wide. The dimensions of page are such that the least possible amount of paper is used.
Thus the width of the page is( ) inches, its height is ( )inches, its total area is ( ) square inches. There's a whole lot of white space on that page, its minimum area not withstanding!

2. Originally Posted by thebristolsound
A rectangular page is designed to contain 72 square inches of print. The margins at the top and bottom of the page are each 4 inches deep. The margins on each side are 2 inches wide. The dimensions of page are such that the least possible amount of paper is used.
Thus the width of the page is( ) inches, its height is ( )inches, its total area is ( ) square inches. There's a whole lot of white space on that page, its minimum area not withstanding!
1. Draw a rough sketch.

2. If the printed area is determined by the width x and the length y the area of the sheet of paper is:

$A = (x+4)(y+8)$

3. Additional condition: $x \cdot y = 72$
Solve this equation for y and plug in the term into the first equation. You'll get the function:

$A(x)=\left(\dfrac{72}x+8\right)(x+4)$

4. Differentiate A wrt x and solve the equation A'(x) = 0 for x.

I'll leave the rest for you.

3. Originally Posted by thebristolsound
Can use Algebra, but I'm told is straight forward as a Calculus Problem:

A rectangular page is designed to contain 72 square inches of print. The margins at the top and bottom of the page are each 4 inches deep. The margins on each side are 2 inches wide. The dimensions of page are such that the least possible amount of paper is used.
Thus the width of the page is( ) inches, its height is ( )inches, its total area is ( ) square inches. There's a whole lot of white space on that page, its minimum area not withstanding!
Let x be the width of the page and y be the height. Then we're given that (x - 4)(y - 8) = 72, so y = 72/(x - 4) + 8. Hence the area is xy = 72x/(x - 4) + 8x. Take the derivative of this quantity with respect to x and set it equal to zero to find its minimum value for x > 4. Then use this info to calculate y and the total area xy.