Question

An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of the material will be least when the depth of the tank is half of its width.

Answer 1

Hint: Since we have a square base, so let us assume that the side of the base is x, and the height of the container is y. Find the volume (V) and total surface area (S) of the container by using the formula: $V=lbh$ and $S=2\left( lb+bh+lh \right)$. Since it is mentioned that the tank will hold a given quantity of water, so the volume will be constant. So, if we need to find the least cost of material, we need to minimize the surface area. So, find the derivative of total surface area with respect to x, and the maximum or minimum value of surface area lies at \[\dfrac{dS}{dx}=0\]
Now, to check whether \[\dfrac{dS}{dx}=0\] is maxima or minima, differentiate $\dfrac{dS}{dx}$ with respect to x. If \[\dfrac{{{d}^{2}}S}{d{{x}^{2}}}>0\], it is minima, else it is maxima. Then, find the relation between x and y.

Complete step by step answer:
As we have assumed that the side of the base is x and the height of the container is y.
So, by using the formula: $V=lbh$ and $S=2\left( lb+bh+lh \right)$, we have:
The volume of container:
 \[\begin{align}
  & V=x\times x\times y \\
 & ={{x}^{2}}y......(1)
\end{align}\]
The total surface area of container:
$\begin{align}
  & S=2\left( x\times x+x\times y+x\times y \right) \\
 & =2{{x}^{2}}+4xy......(2)
\end{align}$
Since it is open thank, so the upper surface area would be subtracted from the total surface area, we have:
$S={{x}^{2}}+4xy......(3)$
We can write surface area as: $S={{x}^{2}}+\dfrac{4V}{x}......(4)$
Since the volume of the container is constant. So, if we need to find the least cost of material, we need to minimize the surface area.
So, to find the derivative of total surface area with respect to x, we get:
$\dfrac{dS}{dx}=2x-\dfrac{4V}{{{x}^{2}}}......(5)$
To find maxima or minima value of the surface area, put \[\dfrac{dS}{dx}=0\], we get:
$\begin{align}
  & \Rightarrow 2x-\dfrac{4V}{{{x}^{2}}}=0 \\
 & \Rightarrow 2x=\dfrac{4V}{{{x}^{2}}} \\
 & \Rightarrow 2{{x}^{3}}=4V......(6) \\
\end{align}$
Since we have: $V={{x}^{2}}y$, so we can write equation (6) as:
$\begin{align}
  & \Rightarrow 2{{x}^{3}}=4{{x}^{2}}y \\
 & \Rightarrow x=2y \\
\end{align}$
So, we have depth as half of the width. But we need to confirm that this is the minimum value. So, differentiate $\dfrac{dS}{dx}$ with respect to x and check for various conditions.
Differentiating equation (5) with respect to x, we get:
$\dfrac{{{d}^{2}}S}{d{{x}^{2}}}=2+\dfrac{8V}{{{x}^{3}}}......(7)$
Clearly, \[\dfrac{{{d}^{2}}S}{d{{x}^{2}}}>0\] for any value of x, we can say that, the function is the minimum value of surface area.
Hence proved that the cost of the material will be least when the depth of the tank is half of its width.

Note:
Since it is mentioned in the question that the container has a square base, so we took the length and width of the container as x and height as h. The students must use the formula for the surface area as $S=2\left( lb+bh+lh \right)$ and not as $S=\left( lb+bh+lh \right)$ by mistake. Some students stop after finding the first derivative and conclude their answer. But, that is not the correct way to proceed. They must perform the second derivative test and make sure that the function is a minimum.