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I am currently getting my Masters in Economics. I did not get any exposure to optimization with inequality constraints in my undergrad. I would like to ensure that I am doing this problem correctly. Note: I understand that Khun-Tucker is not necessary for solving this following problem. That being said, it is important that I master this technique, so I want all the practice I can get:

Problem: A firm has two plants with cost functions $c_{1}(y_{1})=4\sqrt{y_{1}}$ and $c_{2}(y_{2})=2\sqrt{y_{2}}$. What is the cost of producing $y=y_{1}+y_{2}$? Assume that $y_{1},y_{2} \geq 0$

I set up the Lagrange in the following way:

$$\mathcal{L}=4\sqrt{y_{1}}+2\sqrt{y_{2}}-\lambda(y_{1}+y_{2}-y)-\mu_{1}y_{1}-\mu_{2}y_{2}$$

The FOC: $$\mathcal{L}_{y_{1}}=\frac{2}{\sqrt{y_{1}}}-\lambda-\mu_{1}=0$$ $$\mathcal{L}_{y_{2}}=\frac{1}{\sqrt{y_{2}}}-\lambda-\mu_{2}=0$$ $$\mathcal{L}_{\lambda}=0 \Longrightarrow y_{1}+y_{2}=y$$ $$\mu_{1}y_{1}=0$$ $$\mu_{2}y_{2}=0$$

My complementary slackness conditions are: $$\mu_{1} \geq 0; \ \mu_{1}=0 \ if \ y_{1}>0$$ $$\mu_{2} \geq 0; \ \mu_{2}=0 \ if \ y_{2}>0$$

I first check for an interior solution. That is, I look to see if there is a solution where $y_{1},y_{2}>0$. This means that $\mu_{1}=\mu_{2}=0$. This means that $$\frac{2}{\sqrt{y_{1}}}=\lambda \ \ and \ \ \frac{1}{\sqrt{y_{2}}}=\lambda$$ I can take the ratio to find that $$\frac{2 \sqrt{y_{2}}}{\sqrt{y_{1}}}=1$$ Using the fact that $y_{1}+y_{2}=y$, we can see that we have the following solution $$y_{1}=\frac{4}{5}y \ \ and \ \ y_{2}=\frac{1}{5}y$$

I now check the second order conditions. We have $C(y)=4\sqrt{y_{1}}+2\sqrt{y_{2}}$ which means $$C_{1}=\frac{2}{\sqrt{y_{1}}}, \ \ C_{2}=\frac{1}{\sqrt{y_{2}}}$$ $$\Longrightarrow C_{11}=-\frac{1}{y_{1}^{3/2}}, \ \ C_{22}=-\frac{1}{2y_{2}^{3/2}}, \ \ C_{12}=C_{21}=0$$ Plugging in the optimal values into the hessian give us:

$$H=\begin{bmatrix} -\frac{5\sqrt{5}}{8y^{3/2}} & 0 \\ 0 & -\frac{5\sqrt{5}}{2y^{3/2}} \end{bmatrix}$$

Now this is where things get messy for me. I know that if the Hessian is semi positive definite, then we have a minimum. If we have that the hessian is semi negative definite, then we have a maximum. This Hessian doesn't appear to be either. I believe this indicates to us that we have a saddle point.

That is, using principal minors, we have:

$$\mid H \mid ^{(1)}<0, \ \ and \ \ \mid H \mid ^{(2)}=|H|>0$$

That is, $$|H|^{(1)}=det(-\frac{5\sqrt{5}}{8y^{3/2}})=-\frac{5\sqrt{5}}{8y^{3/2}}<0$$

$$|H|^{(2)}=|H|=det(\begin{bmatrix} -\frac{5\sqrt{5}}{8y^{3/2}} & 0 \\ 0 & -\frac{5\sqrt{5}}{2y^{3/2}} \end{bmatrix})=(-\frac{5\sqrt{5}}{8y^{3/2}})(-\frac{5\sqrt{5}}{2y^{3/2}})-0=\frac{125}{16y^{3}}>0$$

That being said, I'm pretty sure we are supposed to find that the interior solution is a maximum. I believe I have done something wrong, but I am not sure what.

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$$H=\begin{bmatrix} -\frac{5\sqrt{5}}{8y^{3/2}} & 0 \\ 0 & -\frac{5\sqrt{5}}{2y^{3/2}} \end{bmatrix}$$ Now this is where things get messy for me. I know that if the Hessian is semi positive definite, then we have a minimum. If we have that the hessian is semi negative definite, then we have a maximum. This Hessian doesn't appear to be either. I believe this indicates to us that we have a saddle point.

The Hessian above has two eigenvalues (denoted by $\lambda$): $$ \lambda_1 = -\frac{5\sqrt{5}}{8y^{3/2}} \ \text{ and } \ \lambda_2 = -\frac{5\sqrt{5}}{2y^{3/2}} $$ The expression $y^{3/2}$ is defined only when $y \geq 0$. This implies that both eigenvalues are negative or undefined. Thus the Hessian does not appear to be indefinite.

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  • $\begingroup$ Thank you, this was a rookie mistake. This next question is more of a math question, why am I finding that |H|^(1)<0 and |H|^(2)=|H|>0 (ie. since the upper left entry is negative and the lower right entry is negative). Wouldn't the determinant be positive? $\endgroup$ Oct 20 at 16:08
  • $\begingroup$ What exactly are you denoting by $|H|^{(1)}<0$? $\endgroup$
    – Giskard
    Oct 20 at 16:10
  • $\begingroup$ @MistahWhite Perhaps you should edit your calculations into the body of the question, as otherwise it may be nigh-on-impossible to determine why you are reaching erroneus results. $\endgroup$
    – Giskard
    Oct 20 at 16:11
  • $\begingroup$ Thank you for your suggestion. I just edited the calculations into the question. $\endgroup$ Oct 20 at 16:18
  • $\begingroup$ @MistahWhite I think you may have misread Sylvester's criterion? Or why is it that you think this contradicts negative definiteness? $\endgroup$
    – Giskard
    Oct 20 at 16:21

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