Question

How do you find the cost of materials for the cheapest such container given a rectangular storage container with an open top is to have a volume of $10{m^3}$ and the length of its base is twice the width, and the base costs $\$10$ per square meter and the material for the sides costs $\$6$ per square meter?

Answer 1

Hint: In this question, we want to find the cheapest cost of materials for a rectangular storage container. The shape of the container is cuboid. Here, the relationship between length, width, and height is given. Apply the formula for the area of the rectangular base and lateral surface area, we can find the total cost. If we want to find the cheapest cost, we have to minimize the cost function that is to find the derivative of a cost function.

Complete step by step solution:
Let us assume that the length of the rectangular storage container is x meter, the width of the rectangular storage container is y meter, and the height of the rectangular storage container is z meter.
 In this question, given that the length of the container’s base is twice the width:
Here, the length is x m and the width is y m.
Therefore, we can write,
$ \Rightarrow x = 2y$ ...(1)
In this question, the volume of the container is given that is $10{m^3}$.
We already know the formula of the volume of a cuboid.
Volume of cuboid$ = l \times b \times h$
Substitute all the values.
$ \Rightarrow 10 = x \times y \times z$
Now, substitute the value of x from equation (1).
The value of x is 2y.
$ \Rightarrow 10 = 2y \times y \times z$
That is equal to,
$ \Rightarrow 10 = 2{y^2} \times z$
Let us divide both sides by $2{y^2}$.
$ \Rightarrow \dfrac{{10}}{{2{y^2}}} = \dfrac{{2{y^2} \times z}}{{2{y^2}}}$
That is equal to,
$ \Rightarrow \dfrac{5}{{{y^2}}} = z$ ...(2)
Now, the important thing is to express all three variables in terms of one variable. Here, we will find the area of the rectangular base and the lateral surface area in terms of one variable.
Now, we will find the area of the rectangular base.
The area of the rectangle is $l \times b$.
Therefore,
Area=$l \times b$
Put the values of length and width.
$ \Rightarrow Area = x \times y$
Let us put the value of x to convert the above equation in terms of y.
Put the value of x from the equation (1).
 The value of x is 2y in the above equation.
$ \Rightarrow Area = 2y \times y$
Let us apply multiplication on the right-hand side.
$ \Rightarrow Area = 2{y^2}$ ...(3)
Now, the formula of the lateral surface area is $2\left( {x + y} \right)z$
$ \Rightarrow LSA = 2\left( {x + y} \right)z$
Let us put the value of x and z to convert the above equation in terms of y.
Put the values from the equations (1) and (2).
Therefore,
$ \Rightarrow LSA = 2\left( {2y + y} \right)\dfrac{5}{{{y^2}}}$
That is equal to,
$ \Rightarrow LSA = 2\left( {3y} \right)\dfrac{5}{{{y^2}}}$
$ \Rightarrow LSA = 6y \times \dfrac{5}{{{y^2}}}$
That is equal to,
$ \Rightarrow LSA = \dfrac{{30}}{y}$
Here, the base costs $ \$ 10 $ per sq. m and the lateral surface costs Rs. 6 per sq. m.
So, the total cost is,
$ \Rightarrow total\cos t = 10\left( {area} \right) + 6\left( {LSA} \right)$
Substitute the values of the area and the lateral surface area.
$ \Rightarrow total\cos t = 10\left( {2{y^2}} \right) + 6\left( {\dfrac{{30}}{y}} \right)$
$ \Rightarrow total\cos t = 20{y^2} + \dfrac{{180}}{y}$
So, the function c(y) that we want to minimize the cost is,
$ \Rightarrow c(y) = 20{y^2} + \dfrac{{180}}{y}$
Let us apply the first derivative to the above function. And find the value of y for the function is equal to zero. Apply the formula $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
$c'\left( y \right) = 20\left( {2y} \right) + 180\left( { - 1{y^{ - 2}}} \right)$
That is equal to,
$c'\left( y \right) = 40y - \dfrac{{180}}{{{y^2}}}$
Put the function value zero.
$ \Rightarrow 0 = 40y - \dfrac{{180}}{{{y^2}}}$
Subtract 40y on both sides.
$ \Rightarrow 0 - 40y = 40y - \dfrac{{180}}{{{y^2}}} - 40y$
That is equal to,
$ \Rightarrow - 40y = - \dfrac{{180}}{{{y^2}}}$
Multiply both sides by $ - {y^2}$.
Therefore,
 $ \Rightarrow - 40y\left( { - {y^2}} \right) = - \dfrac{{180}}{{{y^2}}}\left( { - {y^2}} \right)$
That is equal to,
$ \Rightarrow 40{y^3} = 180$
Divide both sides by 40.
$ \Rightarrow \dfrac{{40}}{{40}}{y^3} = \dfrac{{180}}{{40}}$
So,
$ \Rightarrow {y^3} = \dfrac{9}{2}$
Apply cube root on both sides.
$ \Rightarrow y = \sqrt[3]{{\dfrac{9}{2}}}$
This is the cost.

Note:
Here, the important thing is to express all three variables in terms of one variable. Here, we will find the area of the rectangular base and the lateral surface area in terms of one variable. The area of the rectangle is $l \times b$, and the lateral surface area is $2\left( {x + y} \right)z$.