Drum Tight Containers is designing an open-top, square-based, rectangular box that will have a volume of 665.5 in. ${ }^{3}$. What dimensions will minimize surface area? What is the minimum surface area?
What are the dimensions of the box?
The length of one side of the base is $\square \square$
The height of the box is $\square \square$
Answer
Substituting \(x = 11\) and \(h = 5.5\) into the equation for \(A\) gives \(A = 11^2 + 4*11*5.5 = 363\) in^2, which matches our calculated surface area.
Step 1 ：Given that the volume \(V\) of the box is \(665.5\) in^3, we can express \(h\) in terms of \(x\) as \(h = \frac{665.5}{x^2}\).
Step 2 ：Substitute \(h\) into the equation for \(A\), we get \(A = x^2 + 4*x*(\frac{665.5}{x^2}) = x^2 + \frac{2662}{x}\).
Step 3 ：To find the minimum surface area, we need to find the derivative of \(A\) with respect to \(x\) and set it equal to zero.
Step 4 ：The derivative of \(A\) is \(A' = 2x - \frac{2662}{x^2}\).
Step 5 ：Setting \(A'\) equal to zero gives \(2x - \frac{2662}{x^2} = 0\).
Step 6 ：Multiplying through by \(x^2\) gives \(2x^3 - 2662 = 0\).
Step 7 ：Solving for \(x^3\) gives \(x^3 = 1331\).
Step 8 ：Taking the cube root of both sides gives \(x = 11\).
Step 9 ：Substituting \(x = 11\) into the equation for \(h\) gives \(h = \frac{665.5}{11^2} = 5.5\).
Step 10 ：Substituting \(x = 11\) and \(h = 5.5\) into the equation for \(A\) gives \(A = 11^2 + 4*11*5.5 = 121 + 242 = 363\) in^2.
Step 11 ：\(\boxed{\text{So the minimum surface area is 363 in}^2}\).
Step 12 ：Check our results: Substituting \(x = 11\) and \(h = 5.5\) into the equation for \(V\) gives \(V = 11^2*5.5 = 665.5\) in^3, which matches the given volume.
Step 13 ：Substituting \(x = 11\) and \(h = 5.5\) into the equation for \(A\) gives \(A = 11^2 + 4*11*5.5 = 363\) in^2, which matches our calculated surface area.
