p = 2x+y
a = 1/2 y √(x^2 - (y/2)^2)
= 1/2 (p-2x) √(x^2 - ((p-2x)/2)^2)
= 1/4 (p-2x)√(4px-p^2)
da/dx = p(p-3x)/√(4px-p^2)
da/dx=0 when p = 3x
That is, when the triangle is equilateral. I expect you saw that coming, eh?
If an isosceles triangle has a fixed perimeter P, find the dimensions that maximize that area. Show how you know that it is the maximum.
1 answer