SOLUTION: what is the maximum area of a rectangle with a perimeter of 100 feet

Question 391158: what is the maximum area of a rectangle with a perimeter of 100 feet
Found 2 solutions by Alan3354, richard1234:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Max area is a square.
100/4 = 25 foot sides
Area = 25*25 = 625 sq feet

Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Many different ways to solve this.
Solution 1
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If the sides of the rectangle are x and 50-x (so that the perimeter is 100), the area is , which is a parabola in terms of x. The vertex occurs at , or x = 25. It points downward, so x = 25 maximizes the area, so the area is 625.

Solution 2
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Another way is finding the extrema of . We have which is equal to zero when x = 25 (this is actually where the -b/2a rule comes from).
Solution 3
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Yet another way comes from an unusual theorem: AM-GM inequality. Suppose we have two terms and . Then by AM-GM,

Since , ,

The AM-GM inequality says the equality occurs if and only if all the 's are equal, that is, , and the optimal area is 625.

It's pretty rare you'll see a solution like the last one. However AM-GM can be used to prove many similar theorems, i.e. proving that the rectangular solid of fixed surface area that has maximum volume is a cube.
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