Show that among all rectangles of constant perimeter p, the square has the maximum area. {Hint: an optimization problem}

Answer 1

Let P be the perimeter of a rectangle with side lengths x and y, and A the area.

Then P = 2x +2y

→ y = P/2 - x

→ A = x * y = x * (P/2 - x) = Px/2 - x²

1. Note that the graph for this quadratic is a face down parabola, hence will yield a maximum.

→ A' = P/2 - 2x

Setting equal to zero, we get that

2x = P/2

→ x = P/4

→ y = P/2 - P/4 = P/4, making x and y the same length,

making the rectangle a square.

Answer 2

See the videa.