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You know that Area = Length * Width and that Perimeter = 2(Length + Width). Notice that both Area and Perimeter are functions of Length and Width. Anyways, given that Area = 256, we have that 256 = Length * Width. This means that if we know the Length, we would automatically know the Width, since W = 256/L. Now, we want to express perimeter as a function of Length. We know that it's a function of both Length and Width. But just now, we were able to express Width and a function of Length. So Perimeter = 2(Length + 256/Length). Now the perimeter is a function solely of length. Let x represent the Length. Then P(x) = 2(x + 256/x).
There are a couple different solutions to this. The way I'll type here involves calculus. Knowing that you want to minimize the perimeter, that's the same as minimizing the semi-perimeter. So we can look at finding the minimum of f(x) = x + 256/x and whatever x we get that minimizes f(x) will be the same x that minimizes P(x). (You can try minimizing P(x) to see that this is true.) To minimize f(x) from a calculus perspective, you want to first find the derivative of f(x), then find the critical points. f(x) = x + 256/x. So f'(x) = 1 - 256/(x^2). f'(x) is undefined for x = 0, but we don't need to consider that as a possible length which minimizes the perimeter since we want to focus on positive lengths for x. The derivative is 0 whenever 0 = 1 - 256/(x^2). This boils down to x^2 = 256. Or x = 16 or -16. Again, we only need to focus on when x is positive since x represents a length. To see whether x = 16 minimizes f(x), we want to see if f'(x) < 0 for x < 16 and f'(x) > 0 for x > 16. This would give f(x) a local minimum at x = 16. This turns out to be true. And then since x = 16 is the only critical point for f(x) in the interval 0 < x < infinity, it follows that this must be the length we're looking for which minimizes the perimeter. It might not be surprising that x = 16. If you think about it, that means that the length is 16. And so the width is 256/16 = 16. So the rectangle with area 256 which minimizes the perimeter is actually the square. You could generalize that. Of all the rectangles with some fixed area, the square is the one with the minimum perimeter. Another way to say that is "Of all the rectangles with some fixed perimeter, the square has the maximum area."