Question

Find the most economical proportions for a tent as in the figure, with no floor.

Short answer

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

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

Step-by-step solution

Given data

The given tent is without floor.

Concept of Partial Differentiation

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:

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

Derive an expression for the tent without floor

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:

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

Area is given as follows:

$A=2x^2\tan \theta +\frac{2\pi x}{\cos \theta }$

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

Now two independent variables to minimize $\frac{\partial A}{\partial x}=0$and $\frac{\partial A}{\partial \theta }=0$.

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

And calculate as follows:

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

Calculation of the area of the tent without floor

Equate the above two equations, obtain:

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

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

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

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

Now for rectangular part (lower part) without floor

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