The Mathematics Education Department
EMT 725, Lisa F. Garey
Click here for the Problem statement
Here is a common calculus problem solved using a novel approach -- the A.M.-G.M. inequality.
Area of rectangle = ab.
Perimeter of rectangle = 2a+2b.
Half-perimeter = a+b.
Let the given perimeter be P. By the A.M.-G.M. inequality, area = 
with equality iff a=b. Hence, the area of the rectangle is maximized when it is a square, and the maximum area is (1/4 of its perimeter)^2.
Let the given area be A. By the A.M.-G.M. inequality, perimeter = 
with equality iff a=b. Hence, the perimeter of the rectangle is minimized when it is a square, and the minimum perimeter is four times the square root of its area.
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