SOLUTION: a rectangle has a perimeter of 100 cm. Find the greatest possible area for the rectangle. so can you explain how to do this and the complete solution in terms of centimeters. Thank

Question 394456: a rectangle has a perimeter of 100 cm. Find the greatest possible area for the rectangle. so can you explain how to do this and the complete solution in terms of centimeters. Thanks RK
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
If x and y are the side lengths, then 2(x+y) = 100 --> x+y = 50. We can write y = 50-x, so we have x and 50-x as the side lengths.

We want to maximize the value of x(50-x) or -x^2 + 50x. Multiple ways to do this:

Solution 1:
Since the function for area is a quadratic in the form , the vertex occurs at -b/2a, or -50/-2 = 25.

Solution 2 (more general solution):
Suppose , then by the Power Rule. The derivative is zero at x = 25, so 25 is a relative extremum. We can check that x = 25 maximizes the value of the function. This is where the -b/2a from the previous solution comes from.

Solution 3 (pretty interesting solution, but rarely used):
By the AM-GM inequality, if x and y are the dimensions such that x+y = 50,

The equality case of AM-GM occurs when x = y, hence x = y = 25.

In any of the cases, the maximum area occurs when the figure is a square, and the area is 25^2, or 625 sq cm.
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