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.

  • 2
    $\begingroup$ So do you mean that the material has no thickness? $\endgroup$
    – WhatsUp
    Commented Mar 18, 2020 at 22:11
  • $\begingroup$ there is no thickness to the material as far as I can tell from the question. $\endgroup$
    – Matt
    Commented Mar 18, 2020 at 22:12
  • $\begingroup$ All related questions I can find are either open-topped or only are about cost $\endgroup$
    – Matt
    Commented Mar 18, 2020 at 22:35
  • $\begingroup$ What does the function $f$ represent, and how did you arrive at it? $\endgroup$
    – saulspatz
    Commented Mar 18, 2020 at 22:40
  • $\begingroup$ 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. $\endgroup$
    – Matt
    Commented Mar 18, 2020 at 22:48

1 Answer 1



  • 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.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .