An open-top rectangular box is being constructed to hold a volume of $150 \mathrm{in}^{3}$. The base of the box is made from a material costing 6 cents $/ \mathrm{in}^{2}$. The front of the box must be decorated, and will cost 9 cents/ in ${ }^{2}$. The remainder of the sides will cost 4 cents $/$ in $^{2}$.
Find the dimensions that will minimize the cost of constructing this box.
Front width:
in.
Depth:
Height:
in.

Step 1 ：Let the width of the front (and back) be denoted as x, the depth as y, and the height as z. The volume constraint gives us the equation \(x*y*z = 150\).

Step 2 ：The cost function to minimize is \(C = 6*x*y + 9*x*z + 4*(2*y*z)\), where the first term is the cost of the base, the second term is the cost of the front, and the third term is the cost of the other sides.

Step 3 ：We can express z in terms of x and y using the volume constraint and substitute this into the cost function: \(z = 150/(x*y)\), \(C = 6*x*y + 1350/y + 1200/x\).

Step 4 ：Taking the derivative of the cost function with respect to x and y, we get \(C_x = 6*y - 1200/x^2\) and \(C_y = 6*x - 1350/y^2\).

Step 5 ：Solving these equations, we get three sets of values for x and y: \[(4*75^{1/3}/3, 3*75^{1/3}/2), (-2*75^{1/3}/3 - 2*3^{5/6}*5^{2/3}*I/3, -3*75^{1/3}/4 - 3*3^{5/6}*5^{2/3}*I/4), (-2*75^{1/3}/3 + 2*3^{5/6}*5^{2/3}*I/3, -3*75^{1/3}/4 + 3*3^{5/6}*5^{2/3}*I/4)\].

Step 6 ：However, the dimensions of a box cannot be negative or complex, so we discard the second and third solutions. The first solution gives us the values of x and y that minimize the cost.

Step 7 ：We can substitute these values into the equation for z to find the height of the box: \(z = 75^{1/3}\).

Step 8 ：Final Answer: The dimensions that will minimize the cost of constructing the box are \(\boxed{x = 4*75^{1/3}/3 \text{ in.}, y = 3*75^{1/3}/2 \text{ in.}, z = 75^{1/3} \text{ in.}}\).