Answer to Question #341952 in Calculus for Stella

Question

Find the lengths of the sides of an isosceles triangle with a given perimeter if its area is to be as great as possible.

Answer 1

Let $x, x$ and $y$ will be the sides of a isosceles triangle. If $P=x+x+y,$ then $y=P-2x.$

The area $S$ of the triangle is

$S=\sqrt{p(p-x)(p-x)(p-y)}$

$S=\sqrt{\dfrac{P}{2}(\dfrac{P}{2}-x)(\dfrac{P}{2}-x)(\dfrac{P}{2}-(P-2x))}$

$S(x)=\dfrac{\sqrt{P}}{4}(P-2x)\sqrt{4x-P}, \dfrac{P}{4}<x<\dfrac{P}{2}$

Find the first derivative with respect to $x$

$S'(x)=(\dfrac{\sqrt{P}}{4}(P-2x)\sqrt{4x-P})'$

$=\dfrac{\sqrt{P}}{4}\big(-2\sqrt{4x-P}+\dfrac{4(P-2x)}{2\sqrt{4x-P}}\big)$ $=\dfrac{\sqrt{P}}{2\sqrt{4x-P}}(P-4x+P-2x)$

$=\dfrac{\sqrt{P}(P-3x)}{\sqrt{4x-P}}$

Find the critical number(s)

$S'(x)=0=>\dfrac{\sqrt{P}(P-3x)}{\sqrt{4x-P}}=0$

Since $\dfrac{P}{4}<x<\dfrac{P}{2},$ we have the critical number

$x=\dfrac{P}{3}$

If $\dfrac{P}{4}<x<\dfrac{P}{3}, S'(x)<0,S(x)$ decreases.

If $\dfrac{P}{3}<x<\dfrac{P}{2}, S'(x)>0,S(x)$ increases.

The funcrion $S(x)$ has a local maximum at $x=\dfrac{P}{3}.$

Since the function $S(x)$ has the only extremum for $\dfrac{P}{3}<x<\dfrac{P}{2},$ then

the funcrion $S(x)$ has the absolute maximum for $\dfrac{P}{3}<x<\dfrac{P}{2}$ at $x=\dfrac{P}{3}.$

$x=\dfrac{P}{3}, y=\dfrac{P}{3}$

The equilateral triangle has the maximum area.