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I am having troubles in solving correctly the following problem:

A company wants to minimize its total costs, on the condition that the income obtained from the sale of the quantities $x_1, x_2$ of the two products it produces exceed a certain minimum threshold. Knowing that the unit costs of manufacturing each good are linear functions of the produced outputs in the form $C_1 = x_1, C_2 = 2x_2$, that everything that is produced is sold and that the sale prices of the products are: $p_1 = 1$ and $p_2 = 3$, respectively. Determine the quantities $x_1, x_2$ that minimize the cost of the process.

Solution:

$x_1 = 6/11$

$x_2 = 9/11$

$\lambda = -12/11$

$TotalCost(x_1,x_2) = 18/11$

I tried to solve it through the common way: using Lagrange function with Kuhn-Tucker conditions. However, I cannot reach the correct solution, despite I tried several times. I think I am not building Lagrange function correctly as a consequence of not understanding properly the economical meaning of what the problem want me to solve.

So I would be really greateful if you can help me to understand how to reach the correct solution to this specific problem, knowing that clarifying how to build Lagrange function and its restrictions is probably what it is needed here to fully understand the problem and its solution.

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  • $\begingroup$ Since when Kuhn-Tucker multipliers are negative? Where did you get this exercise? $\endgroup$ – Alecos Papadopoulos Jul 28 '18 at 0:49
  • $\begingroup$ Also, with "income" is it meant "profit" (Revenues minus Costs)? $\endgroup$ – Alecos Papadopoulos Jul 28 '18 at 0:53
  • $\begingroup$ I have always worked with negative lambda and in the rest of cases it worked properly. And according to the handbook, it says "income" (so revenues) and not "profit". Anyway, what I would like to have from you, if possible, is the correct interpretation of the problem, in order to being able to correctly build the Lagrange function. Thank you. $\endgroup$ – Ignacio Valdés Zamudio Jul 28 '18 at 8:11
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The way to solve it is by defining it as a maximization problem. Here is a brief explanation of the scan. enter image description here 1. Formulate the original problem, with $k$ denoting the income (or profit) threshold.

  1. Express it as a maximization problem ($min(f) = - max(-f) $). I also re-stated the income (or profit) condition, although that might be unnecessary after all (see below).

  2. Formulate the lagrangian.

  3. Get the partial derivatives of the lagrangian.

  4. Equate the derivatives to zero and solve for $x_1$ and $x_2$.

  5. The previous step leads to $x_1=2x_2 / 3$.

  6. Substitute $x_1$ and $x_2$ in the income (or profit) condition.

  7. The previous step leads to optimal $x_1$ and $x_2$.

  8. Substitute optimal $x_1$ and $x_2$ in the original problem.

It appears that the problem uses $k=3$ as the minimum threshold for income (or profit), since that threshold gives $TotalCost=18/11$, $x_1=6/11$, and $x_2=9/11$.

I'm unsure how to make sense of negative λ, but (positive) $12/11$ stems from the condition (in step 5) $2x_1=λ$.

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  • $\begingroup$ Thank you for your detailed and clear explanation, Iñaki. I followed everything except why you chose x_1 square and 2x_2 square at the beginning, instead of x_1 and 2x_2 (without square) as it appears it should be, since C1 = x_1 and C2 = 2x_2. I would appreciate if you can clarify the reason of your decision. $\endgroup$ – Ignacio Valdés Zamudio Jul 28 '18 at 11:40
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    $\begingroup$ @IgnacioValdésZamudio The function to optimize is $C_1 x_1 + 2 C_2 x_2$, and the problem indicates that $C_1 = x_1$ and $C_2 = x_2$. Replacing $C_1$ and $C_2$ in the function gives the quadratic terms. $\endgroup$ – Iñaki Viggers Jul 28 '18 at 11:48
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    $\begingroup$ That's right. Finally everything made sense. Thank you very much for your time and explanations. $\endgroup$ – Ignacio Valdés Zamudio Jul 28 '18 at 11:49
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The OP clarified in a comment that a) "income" here means "revenues" and not "profit" (it is not always the case), and b) that Karush-Kuhn-Tucker multipliers work just fine if we define them to be non-positive instead of non-negative (they do, but it is not a widely known fact).

The other bad terminology in the statement of the problem is the "unit cost" one -it is really meant "marginal cost". So we have to obtain the Total Cost function from its partial derivatives. This is easy to do since we see that the cross-partial is zero.

So if

$$\frac {\partial TC}{\partial x_1} = x_1,\;\; \frac {\partial TC}{\partial x_2} = 2x_2$$

it follows that

$$TC = \frac 12 x_1^2 + x_2^2 + FC,\;\;\;FC\geq 0$$

and we want to minimize it subject to the constraint $p_1x_1 + p_2x_2 \geq \bar R$.

PS: It appears that the revenue floor is $3$?

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  • $\begingroup$ Thank you for clarification. And yes, the terminology used in this problem is not the best. But I have two questions here: What does FC correspond to? and Why do you think the revenue floor should be 3? $\endgroup$ – Ignacio Valdés Zamudio Jul 28 '18 at 11:22
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    $\begingroup$ @IgnacioValdésZamudio "FC" is fixed cost. not depending on $x_1,x_2$. It is there for mathematical completeness and correctness, but it does not affect the solution. As for the revenue floor being $3$, by a quick calculation it appears to be consistent with the specific numerical solution you are supposed to find. $\endgroup$ – Alecos Papadopoulos Jul 28 '18 at 16:16
  • $\begingroup$ Oh, yes, I got it. Thank you for both clarifications. $\endgroup$ – Ignacio Valdés Zamudio Jul 28 '18 at 16:30

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