91Ó°ÊÓ

The given tent is without floor.

The process of finding the partial derivative of a function is called partial differentiation. In this process, the partial derivative of a function with respect to one variable is found by keeping the other variable constant.

Partial derivative formula:

$$\begin{gathered} {\mathit{f}}_{\mathbf{x}}\mathbf{=}\frac{\mathbf{∂}\mathbf{f}}{\mathbf{∂}\mathbf{x}} \\ {\mathit{f}}_{\mathbf{x}}\mathbf{=}\underset{\mathbf{h}\mathbf{→}\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{h}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{-}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}}{\mathbf{h}} \end{gathered}$$

In the figure the tent can be divided into two parts

The top part as pup tent and another one is cuboid.

Volume is given as follows:

$$\begin{gathered} V=\frac{1}{2}bhl \\ V=z{x}^2\tan \mathrm{θ} \end{gathered}$$

Area is given as follows:

$A=2x^2\tan \mathrm{θ}+\frac{2\mathrm{π}x}{\cos \mathrm{θ}}$

Since $x,z,\mathrm{θ}$are independent $A=2x\tan \mathrm{θ}+\frac{2V}{x\cos \mathrm{θ}\tan \mathrm{θ}}$.

Now two independent variables to minimize $\frac{∂A}{∂x}=0$and $\frac{∂A}{∂\mathrm{θ}}=0$.

$$\begin{gathered} \frac{∂A}{∂x}=4x\tan \mathrm{θ}-\frac{2V\cos ec\mathrm{θ}}{x^2} \\ \frac{∂A}{∂x}=0 \end{gathered}$$

And calculate as follows:

$$\begin{gathered} \frac{∂A}{∂\mathrm{θ}}=2x^2{\sec }^2\mathrm{θ}-\frac{2V\cos ec\mathrm{θ}\cot \mathrm{θ}}{x} \\ \frac{∂A}{∂\mathrm{θ}}=0 \end{gathered}$$

Equate the above two equations, obtain:

$\begin{array}{rcl}\frac{V\cos ec\mathrm{θ}}{2\tan \mathrm{θ}}& =& \frac{V\cos ec\mathrm{θ}\cot \mathrm{θ}}{{\sec }^2\mathrm{θ}}\\ {\cos }^2\mathrm{θ}& =& \frac{1}{2}\\ \mathrm{θ}& =& \frac{\mathrm{π}}{4}\end{array}$

Hence,$V=X^{2}Z$.

$$\begin{gathered} \frac{∂A}{∂x}=2x \\ \frac{∂A}{∂x}=z\sqrt{2} \end{gathered}$$

The high of the upper part of a tent is$x=\frac{1}{2}$.

Now for rectangular part (lower part) without floor

Area is $A=4xy+4xz+zy$.