Question

How do you find the dimensions of the box that minimize the total cost of materials used if a rectangular milk carton box of width $w$ , length $l$ and height $h$ holds $534$ cubic cm of milk and the sides of the box cost $4$ cents per square cm and the top and bottom cost $8$ cents per square cm?

Answer 1

Hint : Consider width and length equal that minimize the total cost of materials used. Express the height as a function of width. Write the expression for the cost obtained in terms of only the width( $w$ ). Then, Derive the cost with respect to width and set it to zero for determination of critical points.

Complete step-by-step answer :
We are given a volume of milk to be $534$ cubic cm of milk, the sides of the box cost $4$ cents per square cm and the top and bottom costs $8$ cents per square cm.
The dimensions given are: length be ‘ $l$ ’, width be ‘ $w$ ’ and height be ‘ $h$ ’.
Since, we know that varying / different length and width other than being equal can reduce the volume for same total, therefore let’s consider width= length/ $w = l$ ;
A/Q , Volume $= w \times l \times h = 534$
Since, $w = l$ so ${w^2}h = 534$
Express the height in terms of width and we get:
$h = \dfrac{{534}}{{{w^2}}}$
We know that, Total cost $=$ (Cost of sides) $+$ (Cost of top and bottom)
$C = \left( {4 \times \left( {4w \times \dfrac{{534}}{{{w^2}}}} \right) + \left( {8 \times \left( {2{w^2}} \right)} \right)} \right) \\\ = 8544{w^{ - 1}} + 16{w^2} \;$
Differentiating with respect to $w$ and setting it to zero to determine critical points, we get:
$\dfrac{{dC}}{{dw}} = 0$ for critical points.
$- 8544{w^{-2}} + 32w = 0$
Assuming that $w \ne 0$ we can multiply by ${w^2}$ and with some simple numeric division:
$- 276 + {w^3} = 0$
And $w = {\left( {267} \right)^{\dfrac{1}{3}}} = 6.44$ (approx)
Since width and length was considered equal, so length $l = 6.44$ and
$h = \dfrac{{534}}{{{w^2}}} = \dfrac{{534}}{{{{\left( {6.44} \right)}^2}}} = 12.88$ (approx)
There length $= 6.44cm$ , width $= 6.44cm$ and height $= 12.88cm$ (approx)
So, the correct answer is “length $= 6.44cm$ , width $= 6.44cm$ and height $= 12.88cm$ (approx)”.

Note : There can be an error in differentiating if formula is not remembered.
Always assume that width is not equal to zero to get the values of other dimensions.
Set the differentiation to zero for critical points.