slader 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

Dimensions of the poster are:

w = 18 in

Y = 36 in

Step-by-step explanation:

Print area of the rectangular poster is 512 in²

Let call "x" and "y" dimensions for the print area of the poster then

A (p) = Print area of the poster = x*y

512 = x*y ⇒ y = 512/x

Total area of the poster is:

A (t) = (y + 4) * (x + 2)

A (t) = y*x + 2*y + 4*x + 8 And as y = 512/x

Total area of the poster as a function of x is:

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

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

Taking derivatives on both sides of the equation we get:

A' (x) = - 1024/x² + 4

A' (x) = 0 ⇒ - 1024 / x² = - 4 ⇒ 4*x² = 1024

x² = 1024/4 ⇒ x² = 256

x = 16 inches

And y = 512/x ⇒ y = 512/16 ⇒ y = 32 inches

So we found x and y dimensions of the print area, then the dimensions of the poster are:

w = x + 2 ⇒ w = 16 + 2 w = 18 in

Y = y + 4 ⇒ Y = 32 + 4 Y = 36 in