Of all rectangles with a perimeter of 3131, which one has the maximum area? (give the dimensions.)

I'll tell you more: when you fix the perimeter of the rectangle, the one with the maximum area is always the square with that perimeter. Here's the proof.

Given a perimeter 2P, all rectangles with that perimeter have sides x and y such that

The area is the product of the dimensions, so

The maximum of this parabola is found by setting its derivative to zero:

which implies

So, the maximum area is achieved when x=y, i.e. when the rectangle is actually a square.

So, the square with perimeter is 3131 has side length

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363