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 angle BAC=a and AC = x
Then the area by the formula A=.5absin(C) is
A = .5xnsin(a)
By the law of cosines
n²= n² + x²  2nxcos(a)
x² = 2nxcos(a)
x = 2ncos(a)
substitute this into the area formula
A = .5*2ncos(a)nsin(a)
= n²cos(a)sin(a)
Use the trig identity sin(2θ) = 2sin(θ)cos(θ) to rewrite this as
A = .5n²sin(2a)
For fixed n this has maximum value when 2a=90º
A = .5n²
Since a=45º substitute this into the first area equation above
.5xnsin(45º) = .5n²
xn(√(2)/2) = n²
x = n√2

Posted by Jer
on 20110616 15:35:29 