# Equation with Side Conditions

A company has two different types of trucks:

Type A has $20\,\text{m}^3$ space for goods which have to be cooled and $40\,\text{m}^3$ space for goods which do not have to be cooled.

Type B has $30\,\text{m}^3$ space for goods which have to be cooled and $30\,\text{m}^3$ space for goods which do not have to be cooled.

The company has to transport $300\,\text{m}^3$ of cooled and $400\,\text{m}^3$ of uncooled goods. The costs per km for trucks of type A is 30 Euro, the cost per km for type B is 40 Euro.

How many trucks of each type have to be used for achieving minimal total costs?

My ideas:

For the cooled food the equation would be: $20x+30y = 300$

For the uncooled food: $40x + 30y = 400$

Additionally, the total price should be minimized, but how exactly? Thanks!

• I have added some more information to my answer. The solution is $(6,6)$, not $(5,7)$, as far as I can tell. This is a fairly difficult problem for high school. – copper.hat Jul 1 '13 at 22:04

The constraints should be inequalities. It may be OK to 'waste' space if the total cost is less.

The cost per km is $c(a,b) = 30a+40b$. The constraints are $20a+30b\ge300$, $40a+30b\ge400$.

This is a good problem in that it shows a potential 'pitfall' of rounding when solving a discrete problem by means of a continuous approximation. I quote pitfall because from a practical standpoint, a slightly non-optimal solution is not a huge problem.

It is a poor high school problem because, while the mathematics are not involved, it does require a little mathematical 'maturity'.

One can work through the continuous problem in the following way: Note that $c(a,b) = 30a+40b = \frac{7}{6} (20a+30b) + \frac{1}{6} (40a+30b)$, and given the constraints, we see that $c(a,b) \ge \frac{7}{6} (300) + \frac{1}{6} (400) = \frac{1250}{3} = 416 \frac{2}{3}$. Furthermore, by assuming both constraints are active, we get a two variable linear system which has a solution $c(5,\frac{20}{3}) = 416 \frac{2}{3}$. Hence this is a (in fact, the unique solution, but that is not relevant here) solution of the continuous problem.

However, we should not stop here. If $a,b$ are integers, we note that $10 \mid c(a,b)$, and so the above bound shows that a lower bound on cost, restricted to integers, must be $c(a,b) \ge 420$. So, the question is, can we find a solution that has cost 420?

The equation c(a,b) = 420 reduces to $3a+4b = 42$, or $a = 14-\frac{4}{3}b$, so we should look for pairs $(14-\frac{4}{3}b, b)$ that satisfy the constraints. Since $a$ must be an integer, and $\gcd(3,4) = 1$, we must have $b = 3 n$ for some integer $n$. Hence we look for solutions of the form $(14-4n, 3n)$.

The first constraint gives $2(14-4n) +3(3n) \ge 30$, or $n \ge 2$.

The second constraint gives $4(14-4n) +3(3n) \ge 40$, or $n \le \frac{16}{7}$. Since $n$ is an integer, we must have $n \le 2$.

Hence setting $n=2$, we see that $(6,6)$ meets the constraints, hence $(6,6)$ is the minimum cost solution. So, the answer is to use $6$ Type A truck and $6$ Type B truck.

• Ok, and I think I would know how to solve this graphically, but how do I solve such a task analytically? – TestGuest Jul 1 '13 at 16:45
• It is a linear program, which is not a precalculus tool, perhaps the solution is obvious graphically. Let me have a look... – copper.hat Jul 1 '13 at 16:47
• well I am in highschool and in the task it tells us also to solve it analytically..>_> – TestGuest Jul 1 '13 at 16:48
• thanks very much for your efforts. – TestGuest Jul 1 '13 at 23:41
• You are very welcome. You could let your teacher know that it is perhaps more complicated than intended (the analytic part, at least). – copper.hat Jul 2 '13 at 0:14