In isosceles triangle AB=BC=n.

What value of AC warrants the largest area of the triangle ABC?
Solve by:
a) Plane geometry.
b) Trigonometry.
c) Calculus.
d) Any other way is welcome.

call AC  = 2x
With AC as the base of the isosceles triangle the height can be found by the Pythagorean theorem. x²+ h² = n² so h = √(n²-x²)

Area is then .5*2x*√(n²-x²)
A(x) = x√(n²-x²)

dA/dx = √(n²-x²) + x*.5*(n²-x²)^-.5*-2x
= √(n²-x²) -x²/√(n²-x²)
= ((n²-x²)-x²)/√(n²-x²)
=(n²-2x²)/√(n²-x²)

set this equal to zero means

n²-2x² = 0
x² = n²/2
x = n/√2 = n√(2)/2
AC = 2x = n√2

Posted by Jer on 2011-06-16 15:17:30