Find the minimum cost of a rectangular box of volume $150 \mathrm{~cm}^{3}$ whose top and bottom cost 3 cents per $\mathrm{cm}^{2}$ and whose sides cost 8 cents per $\mathrm{cm}^{2}$. Round your answer to nearest whole number cents. \[ \text { Cost }= \] cents.

Step 1 ：Let's denote the length, width, and height of the box as \(l\), \(w\), and \(h\) respectively. The volume of the box is given by \(l \times w \times h = 150 \, cm^3\).

Step 2 ：The cost of the box is determined by the area of its surfaces. The top and bottom of the box have a total area of \(2lw\), and the sides have a total area of \(2lh + 2wh\).

Step 3 ：Therefore, the total cost of the box is given by \(3 \times 2lw + 8 \times (2lh + 2wh) = 6lw + 16lh + 16wh\).

Step 4 ：We can express \(h\) in terms of \(l\) and \(w\) using the volume equation: \(h = \frac{150}{lw}\).

Step 5 ：Substitute \(h\) into the cost equation, we get \(Cost = 6lw + 16l\frac{150}{lw} + 16w\frac{150}{lw} = 6lw + \frac{2400l}{w} + \frac{2400w}{l}\).

Step 6 ：To minimize the cost, we take the derivative of the cost with respect to \(l\) and \(w\), set them to zero, and solve for \(l\) and \(w\).

Step 7 ：The derivative of the cost with respect to \(l\) is \(\frac{d(Cost)}{dl} = 6w - \frac{2400}{w} + \frac{2400w}{l^2}\). Setting this to zero gives \(6w^2 - 2400 + \frac{2400w}{l} = 0\).

Step 8 ：The derivative of the cost with respect to \(w\) is \(\frac{d(Cost)}{dw} = 6l - \frac{2400}{l} + \frac{2400l}{w^2}\). Setting this to zero gives \(6l^2 - 2400 + \frac{2400l}{w} = 0\).

Step 9 ：Solving these two equations, we find that \(l = w = 5\sqrt{6} cm\).

Step 10 ：Substitute \(l = w = 5\sqrt{6} cm\) into the volume equation, we get \(h = \frac{150}{(5\sqrt{6})^2} = 5\sqrt{6} cm\).

Step 11 ：Substitute \(l = w = h = 5\sqrt{6} cm\) into the cost equation, we get \(Cost = 6(5\sqrt{6})^2 + \frac{2400(5\sqrt{6})}{5\sqrt{6}} + \frac{2400(5\sqrt{6})}{5\sqrt{6}} = 900 cents\).

Step 12 ：Therefore, the minimum cost of the box is \(\boxed{900}\) cents.