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

0.69 yards

step-by-step explanation:

there are 36 inches in a yard

25 : 36 = 0.69 yards (you can round to 0.70)