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 2011-06-16 15:35:29