91Ó°ÊÓ

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.

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

Now, invoke the Lagrange multiplier method.

$F=2s^{2}cos\left(\mathrm{θ}\right)sin\left(\mathrm{θ}\right)+2s/cos\left(\mathrm{θ}\right)+\mathrm{λ}(2l+4s)$

Take the partial derivative with respect to $\mathrm{θ}$,s and I.

$$\begin{gathered} \frac{∂F}{∂\mathrm{θ}}=2s^2\cos \left(2\mathrm{θ}\right)-2s/\sin \left(\mathrm{θ}\right) \\ =0 \\ \frac{∂F}{∂\mathrm{θ}}=4s\cos \left(\mathrm{θ}\right)\sin \left(\mathrm{θ}\right)+2I\cos \left(\mathrm{θ}\right)+4\mathrm{λ} \\ =0 \end{gathered}$$

Calculate further as follows:

$$\begin{gathered} \frac{∂F}{∂I}=2s\cos \left(\mathrm{θ}\right)+2\mathrm{λ} \\ \frac{∂F}{∂I}=0 \end{gathered}$$

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

Now, with the help of equation (4),

$\mathrm{λ}=-s\cos \mathrm{θ}$

Substitute for $\mathrm{λ}$in equation (3).

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

Now substitute in equation (2).

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

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

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