A rectangular poster is to contain 512 square inches of print. The margins at the top and bottom of the poster are to be 2 inches, and the margins on the left and right are to be 1 inch. What should the dimensions of the poster be so that the least amount of poster is used

Poster dimensions:

w = 18 in

h = 36 in

A (min) = 648 in²

Step-by-step explanation:

Printed area for the rectangular poster is:

512 in²

If we call "x" and "y" dimensions of printed area we have:

512 = x*y ⇒ y = 512/x

According to problem statment dimensions of the poster will be:

w = x + 2 and h = y + 4

Then area of the poster is:

A (p) = w*h ⇒ A (p) = (x + 2) * (y + 4)

A (p) = x*y + 4*x + 2*y + 8

And as y = 512/x we can express A (p) as a function of x

A (x) = x * (512/x) + 4*x + 2 * (512/x) + 8

A (x) = 512 + 4*x + 1024/x + 8 ⇒ A (x) = 520 + 4*x + 1024/x (1)

Taking derivatives on both sides of the equation (1) we get:

A' (x) = 4 - 1024/x²

A' (x) = 0 ⇒ 4 - 1024/x² = 0

4*x² = 1024

x² = 1024/4

x² = 256

x = 16 in and y = 512/16 y = 32 in

The value of A' (16) = 0 since 1024 / 256 = 4

And A'' (x) = - [ - 1024 (2x) / x⁴] will be A'' (x) > 0

Then the function has a minimum for x = 16

And dimensions of the poster are:

w = x + 2 w = 18 in

h = y + 4 y = 36 in

A (min) = 36*18

A (min) = 648 in²