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Conditions:

  • One unit of ski set A takes $6$ hours to process and $1$ hour to varnish.
  • One unit of ski set B takes $4$ hours to process and also $1$ hour to varnish.
  • The workers can only use $1080$ hours for processing and $240$ hours for varnishing in a year.
  • The profit for one unit of A sold is \$$400$, while for B it is \$$300$.

Question:

Assuming all units are sold, how much of each set should be produced to maximize profits in a year?

My reasoning:

The first thing that pops into my head is: a quadratic function with a maxima point that holds the value of total profits. But I have no idea how to actually define the variables in such a way that it would yield a quadratic function.

I've tried to solve the system:

$$\left\{ \begin{array}{c} 6A+4B=1080 \\ A+B=240 \end{array} \right. $$

But this just yields the amount of each set needed to be produced so no hours are wasted, I'm not sure if this would give the most profits however.

P.S. I'm clueless as to what tags I should use, so edits would be appreciated.

Nikolaj Pedersen
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ShootinLemons
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    So you have an Linear optimization problem: $\max 400A+300B$ such that $6A+4B\leq1080$ and $A+B\leq 240$, where $A,B\geq 0$. – Nikolaj Pedersen Dec 21 '21 at 12:05
  • And probably $A,B$ should both be integers. – peterwhy Dec 21 '21 at 12:14
  • Of course, but luckily it is a very well posed question, so we can avoid integer optimization. This kind of nice problem could indicate they only have basic mathematical tools to work with. – Nikolaj Pedersen Dec 21 '21 at 12:19
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    @ShootinLemons What do you know about Linear-programming? What methods do you know, or are supposed to know, to solve this problem. It is a fairly easy problem so it is OK to say you dont know alot. I am just trying to figure out how to help you towards an answer. – Nikolaj Pedersen Dec 21 '21 at 13:10
  • @NikolajPedersen First time I'm hearing about it. I'm fairly experienced in linear subjects however. My understanding of math is High School AP level; calculus and all that, where we covered more than just linear subjects. This specific problem though feels like uncharted waters, really strange. – ShootinLemons Dec 21 '21 at 13:54
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    @ShootinLemons There have been posted an answer now, but the problem is simple enough, that it was most likely meant as an opportunity for you to discover the basic premises of Linear-programming for yourself. Linear-programming can feel like a very different approach compared to calculus, but it can be more intuitive to work with. – Nikolaj Pedersen Dec 21 '21 at 14:36
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    @ShootinLemons If you want to apply the graphical method, see here (first example) how it works. – callculus42 Dec 21 '21 at 15:42

1 Answers1

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Your problem is a linear programming problem, with a small twist. Here, I am giving a geometric solution. Suppose $ x $ units of ski set A and $ y $ units of ski set B are produced. The objective is to find

$\begin{align*} \arg \max_{x, y} (400x + 300y) \;\; \text{such that} \;\; (6x + 4y) \le 1080 ,\; x + y \le 240 \;\text{and} \; x, y \ge 0 \text{ are integers}. \end{align*}$

Lets draw the feasible set of $ (x, y) $: enter image description here

In the figure above, the feasible set is the yellow colored region, which is bounded by the lines $ 6x + 4y = 1080 $, $ x + y = 240 $, $ x = 0 $ and $ y = 0 $. This is called the the feasible set because the $ (x, y) $ points lying in this region satisfy $ (6x + 4y) \le 1080 $, $ x + y \le 240 $ and $ x, y \ge 0 $. Evidently, within this region, we have to find an integer pair $ (x, y) $ such that $ (400x + 300y) $ is maximum.

Now, this region is a polygon, and a line of the form $ 400x + 300y = c $, where $ c $ is any constant, is not parallel to either of the lines $ (6x + 4y) = 1080 $, $ x + y = 240 $, $ x = 0 $ and $ y = 0 $, which form the boundaries of the polygon. Hence, the maximum value of $ c $ in $ 400x + 300y = c $ will be attained when $ (x, y) $ is any of the vertices of the polygon. This can also be seen in the figure, where several lines of the form $ 400x + 300y = c $ are drawn (presented as dashed red lines). It is seen that the vertex $ (x, y) = (60, 180) $ at the intersection of the lines $ (6x + 4y) = 1080 $, $ x + y = 240 $ maximizes $ 400x + 300y = c $, and the maximum $ c $ is obtained to be 78000.

It also happens to be that the maximizer $ (x, y) $ pair is an integer pair $ (60, 180) $, and hence it is the answer to your problem.

That the optimum solution of such a constrained linear optimization problem occurs at the vertices of the feasible set is the central idea of linear programming. However, the difference of your problem with an usual linear programming problem was the requirement of $ x $ and $ y $ to be integers. The formulation of the problem made finding the solution easy, as the optimizing vertex was an integer pair. If that was not so, we would have needed to find the optimum integer pair, which would likely been an integer pair near the vicinity of the optimum vertex.

joy
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