# Thread: Applications of differentiation : Max and Minimization

1. ## Applications of differentiation : Max and Minimization

(1)A rectangular tank with no top and a volume of 8m3 is to be made from a sheet of metal. The width of the base is 4m. Show that the surface area is given by
S = 4x + 4 + 16/x , where x is the length of the base and find the dimensions of the tank so that minimum material is used.

(2) The breaking weight of a cantilever beam is directly proportional to b*d^2, where b is the breadth of the beam and d is its depth. Find the cross-section dimensions of a beam with a cross-section perimeter of 60cm that has maximum breaking weight.

(3)An oil tank, designed to hold 10 m3, is in the form of a cylinder with hemispherical top.
(i) Show that the surface area (A) of the tank is given by A = 20/r + 5/3 * πr^2 , where r is the radius and h is height of the cylinder.
(ii) Find the dimensions (r and h) of the tank that has least surface area and calculate the least surface area.

2. Maxima and minima of differentiable functions are always attained at either boundary points (the largest or smallest value of $x$ possible, for instance) or points at which the tangent line to the curve is horizontal, i.e., $f'(x)\,=\,0$.

(1) To find the surface area $S$ of a tank in terms of its length $x$, we begin by writing down what we know:

\begin{aligned}
wxh&=8\,\mbox{m}^3\\
wx+2wh+2xh&=S.
\end{aligned}

Since $w=4$, we have

\begin{aligned}
4xh&=8\,\mbox{m}^3\\
h&=\frac{2}{x}\,\mbox{m}^3.
\end{aligned}

All that remains is to substitute these expressions in the first equation. To find the value of $x$ that allows us minimum surface area, we look for points at which $S'(x)=0$:

\begin{aligned}
S'(x)&=\frac{d}{dx}\left(4x+\frac{16}{x}+4\right)\ \
&=\frac{d}{dx}(4x)+\frac{d}{dx}\left(\frac{16}{x}\ right)+\frac{d}{dx}(4)\\
&=4-\frac{16}{x^2}+0\\
&=4-\frac{16}{x^2}=0.
\end{aligned}

As we know that $S$ is always positive and that the surface area would tend to $\infty$ as $x$ approached $0$ or $\infty$, we can be sure that the mininum is not attained at a boundary point. All we must do, then, is find the solution of $S'(x)=0$.

(3) Here's a hint:

\begin{aligned}
S&=2\pi rh + \frac{1}{2}\cdot4\pi r^2 + \pi r^2\;\;\;\;\;\;\;\;\;\;\mbox{(cylinder + half sphere + base)}\\
A&=\pi r^2h + \frac{1}{2}\cdot\frac{4}{3}\pi r^3 = 10\,\mbox{m}^3.
\end{aligned}

To find $S$ in terms of $r$, we may begin by finding $h$ in terms of $r$.

Hope this helps!