Arectangular poster is to contain 722 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?

Let
x > the height of the whole poster
y > the width of the whole poster

We need to minimize the area A=x*y

we know that
(x-4)*(y-2)=722
(y-2)=722/(x-4)
(y)=[722/(x-4)]+2

so
A(x)=x*y>A(x)=x*{[722/(x-4)]+2}
Need to minimize this function over x > 4

find the derivative> A1 (x)
A1(x)=2*[8x²-8x-1428]/[(x-4)²]

for A1(x)=0
8x²-8x-1428=0
using a graph tool
gives x=13.87 in

(y)=[722/(x-4)]+2

y=[2x+714]/[x-4]> y=[2*13.87+714]/[13.87-4]> y=75.15 in

the answer is
the dimensions of the poster will be
the height of the whole poster is 13.87 in
the width of the whole poster is 75.15 in