Question

A roof gutter is to be made from a long strip of sheet metal, $\mathbf{24}\mathbf{}\mathbf{\text{cm}}$wide, by bending up equal amounts at each side through equal angles. Find the angle and the dimensions that will make the carrying capacity of the gutter as large as possible.

Short answer

The angle and the dimension that will make the carrying capacity of the gutter as large as possible $\theta ={\cos }^{-1}\frac{4}{3}$.

Step-by-step solution

Given data

A roof gutter is to be made from a long strip of sheet metal $24\text{cm}$wide by bending up equal amounts at each side through equal angles.

Concept of Partial Differentiation

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

The formula to find the Partial derivative is given as:

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

Differentiate the equation made for the capacity of the gutter

Take the equation$A=24l\sin \theta -2l^2\sin \theta +.5*l^2\sin \left(2\theta \right)$ and take the partial derivatives.

Take two equation and set them to 0.

$\begin{array}{rcl}24\sin \theta -4l\sin \theta +2l\sin \theta \cos \theta & =& 0\\ 24l\cos \theta -(2l{)}^{2\cos \theta }+(2l{)}^{2\cos \left(2\theta \right)}& =& 0\end{array}$

Substitute $\frac{{\left(24\right)}^2+\sin \left(\theta \right)}{4\sin \theta -\sin (\theta +2)}-\frac{2*{\left(24\sin \theta \right)}^2}{24\sin \theta -\sin 2\theta }$and get 90 degrees.

$\frac{{\left(24\right)}^2+\sin \left(\theta \right)}{4\sin \theta -\sin (\theta +2)}-\frac{2+{\left(24\sin \theta \right)}^2}{{(24\sin \theta -\sin 2\theta )}^2}\cos \theta +\frac{2{\left(24\sin \theta \right)}^2+\cos \left(2\theta \right)}{\left(4\sin \theta -\sin {\left(2\theta \right)}^2\right.}=0$

Calculation for the angle to make carrying capacity of the gutter

Throw out some common factors and it's a bit easier to look at$\sin \left(\theta \right)(4\sin \theta -\sin (2\theta \left)\right)+2{\sin }^2\theta \cos \theta +2{\sin }^2\theta {\cos }^2\theta$ this assumes $(4\sin \theta -\sin (2\theta ){)}^2\ne 0$

Now note that$2\sin \theta \cos 2\theta =\sin 2\theta$.

Then, calculate:

$\begin{array}{rcl}\left(4{\sin }^2\theta -{\sin }^2\left(\theta \right)\cos \theta \right)+2{\sin }^2\theta \cos \theta +2{\sin }^2\theta {\cos }^2\theta & =& 0\\ \left(4{\sin }^2\theta -{\sin }^2\left(\theta \right)\cos \theta \right)+4{\sin }^2\theta \cos \theta & =& 0\\ \left(4{\sin }^2\theta +3{\sin }^2\left(\theta \right)\cos \theta \right)& =& 0\\ {\sin }^2\theta (4+3\cos \theta )& =& 0\end{array}$

Since, it is known that${\sin }^2\theta \ne 0$.

$\begin{array}{rcl}(4+3\cos \theta )& =& 0\\ \cos \theta & =& \frac{-4}{3}\\ \theta & =& {\cos }^{-1}\frac{4}{3}\end{array}$