I am currently trying to formulate a formula for this problem using Multivariable Calculus.
There is a job of shipping $V$ cubic feet of old steel shavings across the river by barge. To do this, you'll make a box to put the shavings in for the barge shipment. Define:
$x = \textrm{length of the box in feet},$
$y = \textrm{width of the box in feet}, \textrm{and}$
$z = \textrm{height of the box in feet}.$
The material for the box consists of five rectangular pieces. The bottom has area $x \cdot y$; the material for it costs $a$ per square foot. So in dollars, bottom cost $= a \cdot y.$
Another two pieces (the sides) have area $x \cdot z$ each; these cost $b$ per square foot. So in dollars side cost $= 2 b \cdot z.$
And the other two pieces (the ends) have area $y \cdot z$; these cost $c$ per square foot. So in dollars, end cost $= 2 c \cdot y \cdot z.$
The tugboat captain tells us that she will charge $d$ dollars for each trip. You have $V$ cubic feet of shavings; so you'll need $\frac{V}{x y z}$ trips. So in dollars, transportation cost = $\frac{d \cdot V}{x y z}.$
The bottom line is overall cost = bottom cost + side cost + end cost + transportation cost.
What should the ratios bottom cost : side cost : end cost : transportation cost be if the overall cost is at a minimum?
I was trying to use Lagrange multipliers but no constraint equation is given. How can i go about obtaining the ratios for obtaining the formula for minimum cost?
Suppose there are $n$ trips. For minimum cost, the box must be full on every trip (otherwise you could have made the same number of trips with a smaller box). So volume of box = $xyz = V/n$. By the inequality for arithmetic and geometric means, $$(axy + 2bxz + 2cyz)/3 \ge (axy\cdot2bxz\cdot2cyz)^{1/3} = (4abcV^2/n^2)^{1/3}.$$ The cost of the box is minimized if AM = GM, which happens if and only if $axy = 2bxz = 2cyz.$ From this you can work out the ratios of the sides of the box. The total cost of the $n$ trips is $$dn + 3(4abcV^2)^{1/3}n^{-2/3}.$$ If $n$ were a continuous variable, elementary calculus shows that the cost would be a minimum when $n = 2(abcV^2/d^3)^{1/5}.$ Since in practice $n$ has to be an integer, I think you would have to take the integers on each side of the theoretical minimum and see which gives the smaller cost.