A rectangular 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?
Accepted Solution
A:
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