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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.

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    $\begingroup$ 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

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Hints:

  • 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.
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