What proportions will maximize the area shown in the figure (rectangle with isosceles triangles at its ends) if the perimeter is given?

An isosceles triangle.

Two or more independent variables, an unknown function that depends on those variables and partial derivatives of the unknown function with respect to the independent variables make up a partial differential equation.

From the given information, it is a typical optimization problem with constraint, where the perimeter ( p ) is kept constant and we need to find the proportion that will be maximize the area ( A ).

Before proceed, it may anticipate the results, as the maximization of -sided rectangular object leads to symmetric geometry at the end.

So, to prove it will use Lagrange multiplier method.

$$\begin{gathered} p=2I+4s=cons\tan t \\ A=2A_{triangle}+A_{rec\tan gle} \\ A=2\times \frac{1}{2}\left(2s\cos \left(\theta \right)\right)s\sin \left(\theta \right)+I\times 2s\cos \left(\theta \right) \\ A=2s^2\cos \left(\theta \right)\sin \left(\theta \right)+2s/\cos \left(\theta \right) \end{gathered}$$ ……. (1)

Now, invoke the Lagrange multiplier method.

$F=2s^{2}cos\left(\theta \right)sin\left(\theta \right)+2s/cos\left(\theta \right)+\lambda (2l+4s)$

Take the partial derivative with respect to $\theta$,s and I.

$$\begin{gathered} \frac{\partial F}{\partial \theta }=2s^2\cos \left(2\theta \right)-2s/\sin \left(\theta \right) \\ =0 \\ \frac{\partial F}{\partial \theta }=4s\cos \left(\theta \right)\sin \left(\theta \right)+2I\cos \left(\theta \right)+4\lambda \\ =0 \end{gathered}$$

Calculate further as follows:

$$\begin{gathered} \frac{\partial F}{\partial I}=2s\cos \left(\theta \right)+2\lambda \\ \frac{\partial F}{\partial I}=0 \end{gathered}$$

Now by three partial derivative equations and the original constraints equation will be solve for unknown proportions.

Now, with the help of equation (4),

$\lambda =-s\cos \theta$

Substitute for $\lambda$in equation (3).

$$\begin{gathered} 2s\cos \left(\theta \right)\sin \left(\theta \right)+I\cos \left(\theta \right)+2\times \left(-s\cos \left(\theta \right)\right)=0 \\ \cos \left(\theta \right)\left(2s\sin \left(\theta \right)+I+-2s\right)=0 \\ \theta ={\sin }^{-1}\left(\frac{2s-1}{2s}\right) \end{gathered}$$

Now substitute in equation (2).

$$\begin{gathered} s\left(s\cos \left(2\theta \right)-I\sin \left(\theta \right)\right)=0 \\ s\cos \left(2\theta \right)=I\sin \left(\theta \right) \\ s\left(1-2{\sin }^2\left(\theta \right)=I\sin \left(\theta \right)\right) \\ s\left(1-2{\left(\frac{2s-I}{2s}\right)}^2\right)=I\left(\frac{2s-I}{2s}\right) \end{gathered}$$

Similarly, calculate further.

$$\begin{gathered} 2s^2-{\left(2s-I\right)}^2=2sI-I^2 \\ 2s^2+I^2=2sI+{\left(2s-I\right)}^2 \\ 2s^2+I^2=2sI+\left(4s^2-4sI+I^2\right) \\ 2s\left(s-I\right)=0 \end{gathered}$$

So,$s=l,\theta ={\sin }^1\left(\frac{1}{2}\right)={30}^0$.

Thus,$s=l$and $\theta ={30}^{°}$is the criteria of a regular hexagon.