Question

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

Answer 1

We assume three variables as three dimensions of the box. Using the given relations between length and width we calculate the value of height by substituting these values in the formula of volume of cuboid as volume is given as a constant to us. Similarly we assume cost of material for bottom as a variable and calculate cost of material for top and four sides. Form an equation for total cost by multiplying the area of respective edges of the box to cost of material. Form complete equation in one measure (b) and calculate first and second derivative. Use the concept of maxima and minima to find the economic dimensions.

• Volume of a cuboid having length ‘l’, breadth ‘b’ and height ‘h’ is $l \times b \times h$ cubic units.
• Differentiation of${x^n}$ with respect to x is given by $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
• If the second derivative of a function is less than 0 then it has a local maxima and if is greater than 0 then it has local minima.

Step-By-Step answer:
We assume the length, breadth and height of the box as ‘l’, ‘b’ and ‘h’ respectively. Since box is of cuboid shape we can calculate its volume as $l \times b \times h$
Since we are given volume of box is constant, let the constant value be ‘x’
$\Rightarrow x = l \times b \times h$ … (1)
Now we are given that box is twice as long as its width, which means length of the box is two times the breadth
$\Rightarrow l = 2b$
Substitute value of l in equation (1)
$\Rightarrow x = 2b \times b \times h$
Calculate the value of height by shifting all variables except ‘h’ to LHS of the equation
$\Rightarrow \dfrac{x}{{2{b^2}}} = h$ … (2)
Now let us assume the cost of material required for the bottom be Rs. Y per square meter.
We know the bottom of the box has dimensions $l \times b$
$\Rightarrow$Cost for making bottom of box $= Y(l \times b)$ … (3)
Since the material on the top and the four sides costs three times as much per square meter as that in the bottom, we can write
The cost of material required for the top and 4 sides will be Rs. 3Y per square meter
We know the top and 4 sides will have area $(lb + 2bh + 2hl)$
$\Rightarrow$Cost for making top and 4 sides of box $= 3Y(lb + 2bh + 2hl)$ … (4)
Then total cost of making the box is given by sum of equations (3) and (4). Let total cost be represented by T
$\Rightarrow T = Y(lb) + 3Y(lb + 2bh + 2hl)$
Substitute the value of $\dfrac{x}{{2{b^2}}} = h,l = 2b$in RHS of the equation
$\Rightarrow T = Y(2b \times b) + 3Y(2b \times b + 2b \times \dfrac{x}{{2{b^2}}} + 2 \times \dfrac{x}{{2{b^2}}} \times 2b)$
Cancel same factors from numerator and denominator of the fractions
$\Rightarrow T = 2Y{b^2} + 3Y(2{b^2} + \dfrac{x}{b} + \dfrac{{2x}}{b})$
Add fractions with same denominator
$\Rightarrow T = 2Y{b^2} + 3Y(2{b^2} + \dfrac{{3x}}{b})$
Differentiate the equation formed with respect to ‘b’
$\Rightarrow \dfrac{{dT}}{{db}} = \dfrac{d}{{db}}\left( {2Y{b^2} + 3Y(2{b^2} + \dfrac{{3x}}{b})} \right)$
$\Rightarrow \dfrac{{dT}}{{db}} = \left( {4Yb + 3Y(4b - \dfrac{{3x}}{{{b^2}}})} \right)$
Calculate values in RHS by multiplying brackets
$\Rightarrow \dfrac{{dT}}{{db}} = \left( {4Yb + 12Yb - \dfrac{{9xY}}{{{b^2}}}} \right)$
$\Rightarrow \dfrac{{dT}}{{db}} = \left( {16Yb - \dfrac{{9xY}}{{{b^2}}}} \right)$ … (5)
We can calculate maximum or minimum values of ‘b’ by equating the differentiation to 0
$\Rightarrow \dfrac{{dT}}{{db}} = 0$
$\Rightarrow \left( {16Yb - \dfrac{{9xY}}{{{b^2}}}} \right) = 0$
Take LCM in LHS of the equation
$\Rightarrow \left( {\dfrac{{16Y{b^3} - 9xY}}{{{b^2}}}} \right) = 0$
Shift denominator to RHS
$\Rightarrow 16Y{b^3} - 9xY = 0$
Take Y common and cancel the value
$\Rightarrow 16{b^3} - 9x = 0$
Shift 9x to RHS
$\Rightarrow 16{b^3} = 9x$
Divide both sides by 16
$\Rightarrow {b^3} = \dfrac{{9x}}{{16}}$
Take cube root on both sides of the equation
$\Rightarrow {\left( {{b^3}} \right)^{1/3}} = {\left( {\dfrac{{9x}}{{16}}} \right)^{1/3}}$
$\Rightarrow b = {\left( {\dfrac{{9x}}{{16}}} \right)^{1/3}}$ … (6)
Now we again differentiate equation (5) with respect to b
$\Rightarrow \dfrac{{{d^2}T}}{{d{b^2}}} = \dfrac{d}{{db}}\left( {16Yb - \dfrac{{9xY}}{{{b^2}}}} \right)$
$\Rightarrow \dfrac{{{d^2}T}}{{d{b^2}}} = \left( {16Y + \dfrac{{18xY}}{{{b^3}}}} \right)$
Check the sign for double derivative
$\Rightarrow \dfrac{{{d^2}T}}{{d{b^2}}} = \left( {16Y + \dfrac{{18xY}}{{{b^3}}}} \right)$
$\Rightarrow \dfrac{{{d^2}T}}{{d{b^2}}} = Y\left( {16 + \dfrac{{18x}}{{{b^3}}}} \right)$
Shift constant to RHS
Substitute the value of ${b^3} = \dfrac{{9x}}{{16}}$
$\Rightarrow \dfrac{{{d^2}T}}{{d{b^2}}} = Y\left( {16 + \dfrac{{18x}}{{\dfrac{{9x}}{{16}}}}} \right)$
Cancel same factors from numerator and denominator
$\Rightarrow \dfrac{{{d^2}T}}{{d{b^2}}} = Y\left( {16 + 32} \right)$
$\Rightarrow \dfrac{{{d^2}T}}{{d{b^2}}} = 48Y > 0$
Since double derivative is greater than 0, so the cost is minimum when $b = {\left( {\dfrac{{9x}}{{16}}} \right)^{1/3}}$
Substitute value of b to calculate length and height
$\because l = 2b$
$\Rightarrow l = 2{\left( {\dfrac{{9x}}{{16}}} \right)^{1/3}}$
$\because \dfrac{x}{{2{b^2}}} = h$
$\Rightarrow h = \dfrac{x}{{2{{\left( {\dfrac{{9x}}{{16}}} \right)}^{1/3}}}}$
$\Rightarrow h = \dfrac{x}{2}{\left( {\dfrac{{16}}{{9x}}} \right)^{1/3}}$

So, the economical dimensions are $l = 2{\left( {\dfrac{{9x}}{{16}}} \right)^{1/3}}$,$b = {\left( {\dfrac{{9x}}{{16}}} \right)^{1/3}}$ and $h = \dfrac{x}{2}{\left( {\dfrac{{16}}{{9x}}} \right)^{1/3}}$.

Note: Many students make mistake while calculating the first derivative as they write derivative of $\dfrac{1}{x}$ as $\log x$, keep in mind that is the answer for integration, for this you can write $\dfrac{1}{x} = {x^{ - 1}}$ and then differentiate it according to the rule. Also, many students make mistakes when checking maxima and minima. Keep in mind when the value of the double derivative comes positive that means the answer is minimum.