Question

A box of constant volume c is to be twice as long as it is wide. The material on the top and four sides cost three times as much per square meter as that in the bottom. What are the most economic dimensions?

Answer 1

Hint: The correct definition of the variables allows binding the equations. Here, we take the variables $l,b$ and $h$ as the $l = 2b$ dimensions of the box length, breadth and height respectively. With the help of these variables we will define objective function, in the given question it is the final cost of the box.

Complete step-by-step answer:
Let the length, breadth and height of the box be $l,b$and $h$respectively.
According to the question: length of the box is twice its breadth i.e. we can write $l = 2b$.
Volume of the box=\[length \times breadth \times height\]
So, we get
Volume of the box=$2b \times b \times h$
Given volume of the box=c
Hence c=$2{b^2} \times h$
Also, given that area of bottom=area of top=$length \times breadth$
=$2{b^2}$
Area of sides=$2bh + 2bh + bh + bh = 6bh$
If r be the cost of material for bottom, then for top and sides cost will be 3r.
Let us define a function$F$such that,
$\therefore F = r(2{b^2}) + 3r(2{b^2} + 6bh)$
$ \Rightarrow F = r(8{b^2} + 18bh)$…………….(i)
As c=$2{b^2} \times h$
$\therefore \dfrac{{2{b^2}}}{c} = h$
Substituting the value of $h$in (i) we get,
$F = r(8{b^2} + \dfrac{{9c}}{b})$
Now, differentiating the above function with respect to $b$ and equating it to zero we get,
$F = r(8{b^2} + \dfrac{{9c}}{b})$
As $\dfrac{{dF}}{{db}} = 0$
$\therefore r(16b - \dfrac{{9c}}{{{b^2}}}) = 0$
Solving the above equation, we have
$b = {\left( {\dfrac{{9c}}{{16}}} \right)^{\dfrac{1}{3}}}$
Now we will again above function with respect to $b$, we get
$\dfrac{{{d^2}F}}{{dd{b^2}}} = r\left( {16 + \dfrac{{18c}}{{{b^3}}}} \right) > 0$which is positive and hence minimum.
Therefore the most economic dimensions are:
Length=$2{\left( {\dfrac{{9c}}{{16}}} \right)^{\dfrac{1}{3}}}$ ,
Breadth= ${\left( {\dfrac{{9c}}{{16}}} \right)^{\dfrac{1}{3}}}$ and
Height= \[{\left( {\dfrac{{32c}}{{81}}} \right)^{\dfrac{1}{3}}}\].

Note: To find the maximum or minimum values always find the different ion of the given function and put it equal to zero. Here $\dfrac{{{d^2}F}}{{dd{b^2}}} > 0$ is positive and hence minimum. So we can easily find out the economic dimensions. In case we need to find the highest dimensions then $\dfrac{{{d^2}F}}{{dd{b^2}}} < 0$.