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 = √(n²x²)
Area is then .5*2x*√(n²x²)
A(x) = x√(n²x²)
dA/dx = √(n²x²) + x*.5*(n²x²)^.5*2x
= √(n²x²) x²/√(n²x²)
= ((n²x²)x²)/√(n²x²)
=(n²2x²)/√(n²x²)
set this equal to zero means
n²2x² = 0
x² = n²/2
x = n/√2 = n√(2)/2
AC = 2x = n√2

Posted by Jer
on 20110616 15:17:30 