Answer to Question #112257 in Calculus for ryshant

Question

A pentagon is formed by placing an isosceles triangle on a rectangle, as shown in the figure. If the pentagon has fixed perimeter P, find the lengths of the sides of the pentagon that maximize the area of the pentagon

Answer 1

The object is to maximize the area A(x,y,z) of the pictured pentagon while keeping the perimeter P(x,y,z) constant.

$A(x,y,z)=xy+\frac{x}{2}\sqrt{z^2-(\frac{x}{2})^2}$

$A(x,y,z)=xy+\frac{x}{4}\sqrt{4z^2-x^2}$

$P(x,y,z)=x+2y+2z$

$\nabla A=\lambda \nabla P$

$\nabla A=\begin{pmatrix} y+\frac{2z^2-x^2}{2\sqrt{4z^2-x^2}} \\ x \\ \frac{xz}{\sqrt{4z^2-x^2}} \end{pmatrix}$

$\nabla P=\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$

So,we accomplish these equations:

$y+\frac{2z^2-x^2}{2\sqrt{4z^2-x^2}}=\lambda$

$x=2\lambda$

$\frac{xz}{\sqrt{4z^2-x^2}}=2\lambda$

$P=x+2y+2z$

Since P is the only fixed quantity it would be best to solve for x,y and z in terms of P. One way to do that is to first solve for x,y and z in terms of λ then use the fourth equation to solve for λ in terms of P.

This strategy results in:

$x=2\lambda$

$y=(\frac{3+\sqrt{3}}{3})\lambda$

$z=\frac{2\sqrt{3}}{3}\lambda$

$\lambda=\frac{2-\sqrt{3}}{2}P$

Solving for the values of x,y and z in terms of P gives:

$x=(2-\sqrt{3})P$

$y=\frac{3-\sqrt{3}}{6}P$

$z=\frac{2\sqrt{}3-3}{3}P$