# Optimization of a closed box

A box with a rectangular base and top must have a volume of 9m^3. The length of the base is three times the width material for the base costs $5 per square meter. Material for the sides costs $4 per square meter. Material for the top costs $3 per square meter. Find the dimension of the box the minimize the cost. I understand how to solve for an open-top box. The lid is throwing me for a loop. I believe that you can describe the box with equation f(x)=24x^2+(96/x) but am unsure. • So do you mean that the material has no thickness? Mar 18, 2020 at 22:11 • there is no thickness to the material as far as I can tell from the question. – Matt Mar 18, 2020 at 22:12 • All related questions I can find are either open-topped or only are about cost – Matt Mar 18, 2020 at 22:35 • What does the function$f\$ represent, and how did you arrive at it? Mar 18, 2020 at 22:40
• f(x) is being used to describe the cost of the box, therefore, my next step is to try to minimize the cost(f(x)) to get the smallest dimensions. I arrived at it by saying, f(x)=cost=5(Lw)+3(Lw)+8(Lh)+8(hw), then simplifying and combining like terms.
– Matt
Mar 18, 2020 at 22:48

• The solution for a box with a top doesn't differ substantially from that of an open-top box. The area of the top and base are the same—namely $$\ 3w^2\$$ square metres, where $$\ w$$ is the width of the base in metres—so the cost of the top and base is $$\ 15w^2+9w^2=24w^2\$$ as opposed to $$\ 15w^2\$$ for the base alone. So whatever method you used to solve the problem for the open-top box should still work for the box with a top if you simply add $$\ 9w^2\$$ to the cost of the former.
• If the dimensions of the box are $$\ w\times3w\times h\$$, then its volume is $$\ 3w^2h\$$, so if the volume is $$9$$ cubic metres you must have $$\ h=\frac{3}{w^2}\$$. The total area of the two sides with length $$\ w\$$ is $$\ 2\times wh=\frac{6}{w}\$$ and that of the two sides with length $$\ 3w\$$ is $$\ 2\times3wh=\frac{18}{w}\$$. If I multiply the areas of the top, bottom and sides by their costs per unit area and add them, I get $$\ 24w^2+ \frac{96}{w}\$$ for the total cost, corresponding to the equation you believe you can use to describe the box.
• Since the expression for the cost is a function of a single variable $$\ w\$$, the width of the base, you can find the minimum cost by differentiating this function and finding the value of the width for which the derivative is zero.