Isosceles triangle of maximum area with fixed slant side

Text

Among all isosceles triangles with slant side $8$, determine the one of maximum area. (Hint: set the height equal to $x$, the base equal to $y$, and compute the squared area.)

Solution

Let $x$ be the height and $y$ the base. The foot of the height bisects the base, so the Pythagorean theorem on the half-triangle gives $\left(\frac{y}{2}\right)^2+x^2=8^2=64\ \Longrightarrow\ y^2=4(64-x^2).$ The area is $\mathcal A=\dfrac12\,x\,y$, hence $\mathcal A^2=\frac14 x^2 y^2=x^2(64-x^2)=64x^2-x^4.$ Setting $u=x^2$: $\ \mathcal A^2=64u-u^2$, a downward parabola in $u$, maximum at $u=32$. Thus $x=\sqrt{32}=4\sqrt2\approx5{,}66,\qquad y=2\sqrt{32}=8\sqrt2\approx11{,}31,$ $\mathcal A^2=1024\ \Longrightarrow\ \boxed{\mathcal A_{\max}=32.}$ (The optimal triangle is right-angled at the apex: $x=y/2=4\sqrt2$.)