let base be x, and height be y
then c^2 = x^2 + y^2 , where c is a constant
and y = (c^2-x^2)^(1/2)
Area = xy = x(c^2-x^2)^(1/2)
d(Area)/dx = x(1/2)(c^2-x^2)^(-1/2)(-2x) + (c^2 - x^2)(1/2)
= 0 for a max of Area
(1/2)(c^2 - x^2)^(-1/2) [ -2x^2 + 2(c^2-x^2) ] = 0
(c^2 - x^2) [-4x^2 + 2c^2] = 0
x = c ---> not very likely, there would be no rectangle
or
4x^2 = 2c^2
2x^2 = c^2
but c^2 = x^2 + y^2
2x^2 = x^2 + y^2
x^2 = y^2
x = y
so it must be a square, (base = height)
Show that of all rectangles with diagonal length c, the square has the largest area.
1 answer