Courses for Kids
Free study material
Offline Centres
Store Icon

A box with a square base and open top must have a volume of \[32000\;{\text{c}}{{\text{m}}^{\text{3}}}\]. How do you find the dimensions of the box that minimize the amount of material used?

Last updated date: 24th Jul 2024
Total views: 384k
Views today: 5.84k
384k+ views
Hint: In this question, we have the square base box and volume is given. For finding the dimension of the box at minimum amount of material used, we used the two formulas. First is the volume formula and second is the surface area formula.

Complete step by step answer:
In the question, the data is given as below.
The volume of the square base box, which is denoted by the \[V\] is given as.
\[V = 32000\;{\text{c}}{{\text{m}}^{\text{3}}}\]
Let's assume that the length and breadth of a square base box are \[x\;{\text{by}}\;x\] at the top and the height of the square base box is\[h\].
Then the volume of the box is written as below.
\[ \Rightarrow V = {x^2}h\]
Then put the value of volume in the above.
\[ \Rightarrow {x^2}h = 32000\]
\[ \Rightarrow h = \dfrac{{32000}}{{{x^2}}}............\left( 1 \right)\]
After that, we come to the surface area of the square base box.
Let’s, \[S\] is the surface area of the square base box and written as below.
\[ \Rightarrow S = area\;of\;the\;base + total\;area\;of\;the\;4\;sides\]
\[ \Rightarrow S = {x^2} + 4xh\]
Then, put the value of \[h\] from the equation \[\left( 1 \right)\]. Hence the above equation is written as below.
\[ \Rightarrow S = {x^2} + 4x\left( {\dfrac{{32000}}{{{x^2}}}} \right)\]
After calculating, we can write the above equation as below.
  S = {x^2} + 4\left( {\dfrac{{32000}}{x}} \right) \\
   \Rightarrow S = {x^2} + \dfrac{{128000}}{x}........\left( 2 \right) \\
We want to minimize the surface area, and then we write the first derivative.
\[{S^{'}} = 2x - \dfrac{{128000}}{{{x^2}}} = \dfrac{{2{x^3} - 128000}}{{{x^2}}}\]
When, \[{S^{'}} = 0\]
\[ \Rightarrow \dfrac{{2{x^3} - 128000}}{{{x^2}}} = 0\]
We can calculate to find the value\[x\].
  {x^3} = 64000 \\
   \Rightarrow x = 40 \\
Then, we can second derivative of surface area verify for the surface area minimum at the critical value of \[x\].
\[ \Rightarrow {S^{''}} = 2 + \dfrac{{256000}}{{{x^3}}}\]
When we put the value of \[x\] in the second derivative, then the result is positive.
Then, we find the height of the square base box.
Put the value of \[x\] in equation \[\left( 1 \right)\].
  h = \dfrac{{32000}}{{{{\left( {40} \right)}^2}}} \\
  \therefore h = 20 \\

Therefore, the length and breadth of the square base box are \[{\text{40}}\;{\text{cm}}\;{\text{by}}\;{\text{40}}\;{\text{cm}}\] and the height of the square base box is \[{\text{20}}\;{\text{cm}}\] for the minimum amount of material used.

If you want to find the dimensions of a square base box. Then you can use the volume and the surface area formula for calculating the dimension of a square base box. And the formula is given as below.
\[V = {x^2}h\] and \[S = {x^2} + 4xh\].