Asked by o
A poster is to contain 50 square inches of printed matter with margins of 4 inches at the top and bottom and 2 inches at each side. Find the OUTER dimensions if the total area of the poster is the smallest possible.
Answers
Answered by
Steve
xy=50, so y=50/x
a = (x+4)(y+8) = (x+4)(50/x + 8)
a = 8x + 200/x + 82
da/dx = 8 - 200/x^2
da/dx=0 if x=5
so, the inside dimensions are 5x10
and the outside is 9x18
a = (x+4)(y+8) = (x+4)(50/x + 8)
a = 8x + 200/x + 82
da/dx = 8 - 200/x^2
da/dx=0 if x=5
so, the inside dimensions are 5x10
and the outside is 9x18
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