Two sides of a triangle are to be 10 cm long. How long should the third side be to make the triangle have maximum area? (Hint: square the area then differentiate to find the value x for which the area has maximum value)

Answer 1

So the triangle must be isosceles.
Sketch it that way, draw a perpendicular to the missing side.
let that perpendicular, the height, be h cm
let the base be 2x cm (I am avoiding fractions)

A = (1/2)(2x)(h) = xh
then A^2 = x^2h^2, (their hint)

but x^2 + h^2 = 10^2
h^2 = 100 - x^2

A^2 = x^2(100-x^2) = 100x^2 - x^4
d(A^2)/dx = 200x - 4x^3 = 0 for a max/min

200x - 4x^3 = 0
x^3 - 50x = 0
x(x^2 - 50) = 0
x = 0 <----- that would yield the minimum area
or
x^2 = 50
x = 5√2
but I called the base 2x, so the third side must be 10√2 cm

(notice this would make the base angle 45°, and the vertex angle 90°
that is, you have half a square.
This makes sense since the largest area of a rectangle is yielded
when that rectangle is a square )