Question 391158
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 {{{x(50-x) = -x^2 + 50x}}}, which is a parabola in terms of x. The vertex occurs at {{{-b/2a}}}, 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 {{{y = -x^2 + 50x}}}. We have {{{dy/dx = -2x + 50}}} 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 {{{a[1] = x}}} and {{{a[2] = 50-x}}}. Then by AM-GM,


{{{(a[1] + a[2])/2 >= sqrt(a[1]a[2])}}} Since {{{a[1] + a[2] = 50}}}, {{{50/2 = 25}}},


{{{25 >= sqrt(a[1]a[2])}}}


{{{625 >= a[1]a[2]}}}


The AM-GM inequality says the equality occurs if and only if all the {{{a[i]}}}'s are equal, that is, {{{a[1] = a[2] = 25}}}, 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.