Question

A metal box with a square base and vertical height is to contain \[1024\text{ c}{{\text{m}}^{3}}\]. The material for the top and the bottom costs \[Rs.5/c{{m}^{2}}\] and the material for the sides costs \[Rs.2.50/c{{m}^{2}}\]. Find the least cost of the box.

Answer 1

Hint: We have to assume the side of the square base as x and the height of the metal box as y. We have to find the volume of the metal boxes based on the above assumptions. Now, we should equal the volume obtained in variables to \[1024\text{ c}{{\text{m}}^{3}}\]. A relation between x and y can be obtained. In the questions, we were given that The material for the top and the bottom costs \[Rs.5/c{{m}^{2}}\] and the material for the sides costs \[Rs.2.50/c{{m}^{2}}\]. Let us assume the cost of the box as C. Now, we should find the cost of the box in variables. Using the relation between x and y, we can convert C either into x (or) y. To find the minimum cost of the metal box, we should use the maxima and minima concept. A function f(x) is said to be minimum at the value of x where f’(x)=0 and f”(x)>0 and a function f(x) is said to be minimum at the value of x where f’(x)=0 and f”(x)>0. The function f(x) is said to have an inflection point at the value of x where f’(x)=0 and f”(x)=0.

Complete step-by-step solution:
Before solving the question, let us illustrate a metal box with a square base and vertical height as shown in the below figure.

From the question, it is clear that the volume of box is \[1024\text{ c}{{\text{m}}^{3}}\]. Let us assume the side of a square base as x and the height of the box as y.
We know that the volume of any 3D figure is equal to the product of the area of the base of the 3D figure and the height of the 3D figure.
So, the volume of the given metal box is equal to the product of the area of the base of the metal box and the height of the metal box.
Area of square base of metal box = \[{{x}^{2}}\].
Height of metal box = y.
So, the volume of metal box = \[{{x}^{2}}y\].
We know that the volume of box is \[1024\text{ c}{{\text{m}}^{3}}\].
Hence,
\[\begin{align}
  & {{x}^{2}}y=1024 \\
 & \Rightarrow y=\dfrac{1024}{{{x}^{2}}}....(1) \\
\end{align}\]
From the question we have The material for the top and the bottom costs \[Rs.5/c{{m}^{2}}\] and the material for the sides costs \[Rs.2.50/c{{m}^{2}}\].
Area of square base of metal box = \[{{x}^{2}}\]
Area of sides of metal box = 4xy
Total area of metal box = \[2({{x}^{2}})+4(xy)=2{{x}^{2}}+4xy\]
Let us assume the cost of the metal box as C.
\[\begin{align}
  & C=(5)(2{{x}^{2}})+(2.5)(4xy) \\
 & \Rightarrow C=10{{x}^{2}}+10xy \\
 & \Rightarrow C=10x(x+y)......(2) \\
\end{align}\]
Substitute equation (2) in equation (1).
\[\begin{align}
  & \Rightarrow C=10x(x+\dfrac{1024}{{{x}^{2}}}) \\
 & \Rightarrow C=10{{x}^{2}}+\dfrac{10240}{x}.....(3) \\
\end{align}\]
We have to find the minimum cost required to buy the metal box.
We know that a function f(x) is said to be minimum at the value of x where f’(x)=0 and f”(x)>0.
So, to have the cost of metal box as minimum \[\dfrac{dC}{dx}=0\] and \[\dfrac{{{d}^{2}}C}{d{{x}^{2}}}>0\].
Now we should differentiate equation (3).
\[\begin{align}
  & \Rightarrow \dfrac{dC}{dx}=\dfrac{d}{dx}\left( 10{{x}^{2}}+\dfrac{10240}{x} \right)=0 \\
 & \Rightarrow \dfrac{dC}{dx}=\dfrac{d}{dx}(10{{x}^{2}})+\dfrac{d}{dx}\left( \dfrac{10240}{x} \right)=0 \\
 & \Rightarrow \dfrac{dC}{dx}=10(2x)+\dfrac{d}{dx}\left( \dfrac{10240}{x} \right)\text{=0 }\left( \dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}} \right) \\
 & \Rightarrow \dfrac{dC}{dx}=20x-\dfrac{10240}{{{x}^{2}}}=0\text{ } \\
 & \Rightarrow 20x=\dfrac{10240}{{{x}^{2}}} \\
 & \Rightarrow 20{{x}^{3}}=10240 \\
 & \Rightarrow {{x}^{3}}=512 \\
 & \Rightarrow x=8......(4) \\
\end{align}\]
Now let us double differentiate equation (3) on both sides, then we get
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}C}{d{{x}^{2}}}=\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( 10{{x}^{2}}+\dfrac{10240}{x} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}C}{d{{x}^{2}}}=\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( 10{{x}^{2}} \right)+\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( \dfrac{10240}{x} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}C}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 20x \right)-\dfrac{d}{dx}\left( \dfrac{10240}{{{x}^{2}}} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}C}{d{{x}^{2}}}=20-\left( \dfrac{-10240}{{{x}^{3}}} \right)\left( 3 \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}C}{d{{x}^{2}}}=20+\left( \dfrac{30720}{{{x}^{3}}} \right)...(5) \\
\end{align}\]
Now let us substitute equation (4) in equation (5), then we get
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}C}{d{{x}^{2}}}=20+\left( \dfrac{30720}{{{8}^{3}}} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}C}{d{{x}^{2}}}=20+\left( \dfrac{30720}{64} \right)>0 \\
\end{align}\]
So, it is clear that the least cost value of box can be obtained at the value of x is equal to 8.
Now we will substitute equation (4) in equation (3), then we get
\[\begin{align}
  & \Rightarrow C=10{{(8)}^{2}}+\dfrac{10240}{8} \\
 & \Rightarrow C=640+1280 \\
 & \Rightarrow C=1920 \\
\end{align}\]
Hence, the least cost of the box is equal to 1920.

Note: There is a chance of mistake that students may write the total area of box as \[{{x}^{2}}+4xy\] instead of \[2{{x}^{2}}+4xy\]. As we will have to square bases for a metal box, it is required to write the area of the box as \[2{{x}^{2}}+4xy\].
A function f(x) is said to be minimum at the value of x where f’(x)=0 and f”(x)>0 and a function f(x) is said to be maximum at the value of x where f’(x)=0 and f”(x)<0. The function f(x) is said to have an inflection point at the value of x where f’(x)=0 and f”(x)=0.