Given a isosceles triangle with a perimeter of 3m, what is the dimensions that gives the largest area possible?

Question

Given an isosceles triangle with a perimeter of 3m, what are the dimensions that give the largest area possible?

Answer 1

1 m by 1 m by 1 m, all sides = 1

Area = bh/2 = (1)sqr3/2 = sqr3/2 = about 1.732/2 = 0.866 square meter

Let x = the length of 2 sides each

then the 3rd side = 3-2x = base

the height = square root of (x^2 - (3-x)^2/4)

Area = bh/2 = (3-x)(1/2)sqr(4x^2 -9 + 6x -x^2)/2

A = (1/4)(3-x)(3x^2 +6x -9)

=(3/4)(3-x)(3x^2+6x-9)

A'(x) = (9/4)(3-x)(x^2+2x-3)(6x+6) =0

A'(x) = (3-x)(x+1)(x+3)(x-1) = 0

x = 3,-1, -3 and 1 ignore all solutions except x= 1

two sides are 1, third side = 3-2x = 3-2 = 1