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