# Math Help - [SOLVED] Maximization of pentagon area...

1. ## [SOLVED] Maximization of pentagon area...

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

Its how to get the answers that i need help with...

Maximize the area:
$A=xy+xtan(\theta)$
Constraint:
$P=x+2y+x^2+y^2$

using lagrange multipliers:
$=\lambda<1+2x,2+2y>$

$y+tan(\theta)=\lambda(1+2x)$
$x=\lambda(2+2y)$

From here i cannot solve for any one variable. Caused mainly by the $tan(\theta)$, which adds one too many variables. I am most likely missing some little step to get me there, or mabye lagrange multipliers are not the best method to use...Please let me know what to do. Thanks!

2. Originally Posted by snaes
A pentagon is formed by placing an isosceles triangle on a rectange. If the pentagon has fixed perimeter P, find the lengths of the sides of the pentagon that maximize the area of the pentagon.

Its how to get the answers that i need help with...

Maximize the area:
$A=xy+xtan(\theta)$
Constraint:
$P=x+2y+x^2+y^2$

using lagrange multipliers:
$=\lambda<1+2x,2+2y>$

$y+tan(\theta)=\lambda(1+2x)$
$x=\lambda(2+2y)$

From here i cannot solve for any one variable. Caused mainly by the $tan(\theta)$, which adds one too many variables. I am most likely missing some little step to get me there, or mabye lagrange multipliers are not the best method to use...Please let me know what to do. Thanks!
Assuming x is the width of the rectangle, y is the height of the rectangle and h is the height of the isosceles triangle, then A = xy + xh/2, P = x + 2y + 2sqrt((x/2)^2 + h^2) = x + 2y + sqrt(x^2 + 4h^2). Thus grad(A) = <y + h/2, x, x/2> and grad(P) = <1 + x/sqrt(x^2 + 4h^2), 2, 4h/sqrt(x^2 +4h^2)>. Setting grad(A) = lambda * grad(P), then this, together with the constraint on the perimeter, gives a system of four equations in four variables.

3. Ok, it was the tangent $\theta$ that messed me up thanks. Now I got it replaced by h and it works out.