Question 11 (5 points) A rectangular box is to have a volume of $7200 \mathrm{~cm}^{3}$. The length of the box is three times the width. Determine the height of the box with the smallest total surface area. Let $x$ represents the width, $y$ represents the length and h represents the height for this problem. Inelude the diagtim and second derivatiwe lest be full mark.

Step 1 ：Let the width be represented by \(x\), the length by \(y\), and the height by \(h\).

Step 2 ：Since the length is three times the width, we have \(y = 3x\).

Step 3 ：The volume equation is \(V = x * y * h = 7200\).

Step 4 ：Substitute \(y = 3x\) into the volume equation: \(7200 = x * (3x) * h\).

Step 5 ：Solve for \(h\) in terms of \(x\): \(h = \frac{2400}{x^2}\).

Step 6 ：Substitute the expressions for \(y\) and \(h\) into the surface area equation: \(S = 2(x(3x) + x(\frac{2400}{x^2}) + (3x)(\frac{2400}{x^2}))\).

Step 7 ：Simplify the surface area equation: \(S = 2(3x^2 + 2400 + \frac{7200}{x})\).

Step 8 ：Find the first derivative of \(S\) with respect to \(x\): \(S' = 12x - \frac{14400}{x^2}\).

Step 9 ：Find the second derivative of \(S\) with respect to \(x\): \(S'' = 12 + \frac{28800}{x^3}\).

Step 10 ：Find the critical points by setting \(S'\) equal to zero and solving for \(x\): \(x = 2 * 150^{\frac{1}{3}}\).

Step 11 ：Use the second derivative test to confirm that the surface area is minimized at the critical point: \(S''(2 * 150^{\frac{1}{3}}) = 36 > 0\).

Step 12 ：Find the dimensions of the box: width \(x = 2 * 150^{\frac{1}{3}}\), length \(y = 6 * 150^{\frac{1}{3}}\), and height \(h = 4 * 150^{\frac{1}{3}}\).

Step 13 ：\(\boxed{\text{The height of the box with the smallest total surface area is } 4 * 150^{\frac{1}{3}} \text{ cm}}\).