Calculus? just one question?

A rectangular poster is to be made out of cardboard. The margins on the top and the bottom will be 2 inches. The margins on the sides will be 1 inch. the area of the poster devoted to the printed material must be 300 square inches.

(a) To the nearest .001 inches, find the dimensions of the poster that use the least amount of cardboard.

(b) To the nearest .001 inches, find the area of the poster that uses the least amount of cardboard.

Basically you wish to minimize the area of the cardboard used.

The printed area's length I'll call L

The printed area's width will be 300/L

So, the length of the cardboard will be L + 4 to account for 2 inches at top and bottom.

The width of the cardboard will be 300/L + 2

So the Area of the cardboard A = (L+4)(300/L +2)

A = 300 + 2L + 1200/L + 8

A = 308 + 2L +1200(L)^(-1)

A ' = 2 - 1200/(L^2)

0 = 2 - 1200/(L^2)

2 = 1200/(L^2)

2L^2 = 1200

L^2 = 600

L = sqrt(600)

If you calcualte that to the nearest .001 and then substitute back in to find the width of the printed area, you can find the Dimensions of the cardboard and the area of the cardboard.

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