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Out of all the pentagons in the shape of a rectangle overlapped by an isosceles triangle, with perimeter P fixed, determine the dimensions of the one with maximum area.

Maximize area of pentagon

miracle173
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Pedro Gabriel Carvalho
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    Whát hávé yóú trìèd? – Jaideep Khare May 23 '17 at 15:36
  • Have you drawn a diagram and decided upon the variables? – John Wayland Bales May 23 '17 at 15:47
  • @JohnWaylandBales I draw the diagram(sorry, i don't know how to insert photos here), and got the perimeter P = x + 2y + 2z(in which 2x + 2y is the rectangle and 2z + x is the triangle), then i took it apart into 2 areas, the rectangle and the triangle, S(x,y,z) = S1(x,y) + S2(x,z), so my doubt is: to find the maximum area i need to use the gradient vector? – Pedro Gabriel Carvalho May 27 '17 at 01:50
  • OK, I added a diagram to your question. Yes, you should work this by the Lagrange multiplier method which uses the gradients of $P$ and $S$. You want to maximize $S$ subject to the constraint $P$. You will have four equations in the variables $x,y,z,\lambda$. One of the four equations is the equation for $P$, the other three come from $\nabla A=\lambda\nabla P$. – John Wayland Bales May 27 '17 at 05:29

2 Answers2

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

\begin{eqnarray} A(x,y,z)&=&xy+\frac{1}{2}x\sqrt{z^2-\left(\frac{x}{2}\right)^2}\\ &=&xy+\frac{x}{4}\sqrt{4z^2-x^2} \end{eqnarray}

\begin{equation} P(x,y,z)=x+2y+2z \end{equation}

This may be accomplished using the method of LaGrange multipliers, solving

\begin{equation} \nabla A=\lambda\nabla P \end{equation}

\begin{eqnarray} \nabla A(x,y,z) &=&\begin{pmatrix} {y+\dfrac{2z^2-x^2}{2\sqrt{4z^2-x^2}}}\\\\ x\\\\ \dfrac{xz}{\sqrt{4z^2-x^2}} \end{pmatrix}\\ \nabla P(x,y,z)&=&\begin{pmatrix} 1\\2\\2 \end{pmatrix} \end{eqnarray}

So we have

  1. $$ y+\dfrac{2z^2-x^2}{2\sqrt{4z^2-x^2}}=\lambda$$

  2. $$ x=2\lambda $$

  3. $$ \dfrac{xz}{\sqrt{4z^2-x^2}}=2\lambda $$

  4. $$ P=x+2y+2z $$

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 $\lambda$ then use the fourth equation to solve for $\lambda$ in terms of $P$.

This strategy results in

  1. $$x=2\lambda$$

  2. $$y=\left(\dfrac{3+\sqrt{3}}{3}\right)\lambda$$

  3. $$ z=\left(\dfrac{2\sqrt{3}}{3}\right)\lambda $$

  4. $$ \lambda=\left(\dfrac{2-\sqrt{3}}{2}\right)P $$

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

  1. $$x=\left(2-\sqrt{3}\right)P $$

  2. $$ y=\left(\dfrac{3-\sqrt{3}}{6}\right)P$$

  3. $$ z=\left(\dfrac{2\sqrt{3}-3}{3}\right)P$$

I have omitted algebraic details of the last six equations because of the length of time required to render them into MathJax. If OP finds an error in any of these, let me know and I will double-check my steps.

Addendum: I have checked my results twice and found no errors.

There is also a curious feature of the solution.

The following are easily verified:

  1. $xy=\dfrac{9-5\sqrt{3}}{6}P^2$

  2. $x-y=\dfrac{9-5\sqrt{3}}{6}P$

Therefore

$$ \frac{xy}{x-y}=P $$

John Wayland Bales
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Your value of lambda in terms of P does not agree with what I came up with. I got 3/(4*(3+sqrt(3))P - but yours appears to work, so I am not sure what I did wrong.

Steve
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  • Please use mathjax and enclose the expression in between dollar signs '$', '$' . – PNDas Dec 22 '20 at 14:09
  • Just to be sure, I reworked this from the beginning and obtained the same result. For future reference, comments should be made by clicking the "add a comment" link following the answer, not by creating a new answer. – John Wayland Bales Dec 22 '20 at 18:56