Question

A box with a square base and an open top is, to have a volume of 32,000 cm³.Find the dimensions of the box that minimize the amount of material used.

Short answer

40 cm by 40 cm by 20 cm.

Step-by-step solution

Given Data

1)Volume of the box is 32000cm³.

2)The box has a square base and an open top.

 Step 2: Determination of the dimensionsof the box

Let a be the side of the square base and h be the height of the box.

Now, Volume of the box \(V = {a^2}h\)

Minimize the amount of material by minimizing the surface area as the amount of material is directly proportional to the surface area.

The surface area is

\[A = {a^2} + 4ah\]....(1)

Now, \(V = {a^2}h = 32000\;{{\mathop{\rm cm}\nolimits} ^3}\)……….(2)

Eliminate h from above equation.

\(h = \frac{{32000}}{{{a^2}}}\)

Put \(h = \frac{{32000}}{{{a^2}}}\)in equation (1).

\[\begin{array}{c}A = {a^2} + 4a\left( {\frac{{32000}}{{{a^2}}}} \right)\\A = {a^2} + \frac{{128000}}{a}\end{array}\]

Now, to minimize A, differentiate A with respect to a and equate to 0.

\(\begin{array}{c}A' = 0\\2a - \frac{{128000}}{{{a^2}}} = 0\\\frac{{2{a^3} - 128000}}{{{a^2}}} = 0\\{a^3} - 64000 = 0\\{a^3} = 64000\\a = 40\end{array}\)

The critical number is \(a = 40\;{\mathop{\rm cm}\nolimits} \)

Now find second derivative of A to find at what point the function A is critical.

\(A'' = a - \frac{{256000}}{{{a^3}}}\)

Therefore, A is minimum at this critical number \(a = 40\;{\mathop{\rm cm}\nolimits} \).

Now find h by using\(a = 40\;{\mathop{\rm cm}\nolimits} \) and \(h = \frac{{32000}}{{{a^2}}}\).

\(\begin{array}{c}h = \frac{{32000}}{{{a^2}}}\\h = \frac{{32000}}{{{{40}^2}}}\\h = 20\;{\mathop{\rm cm}\nolimits} \end{array}\)

Thus, the box should have 40 cm by 40 cm base and 20 cm height.