<p>What's tricky is that the problem has a kink. Below a certain production value no capital is used. The demand functions for $k$ and $l$ are as follows:</p>

<p>If $Q \leq 180$
$$L(Q) = (Q/20)^2$$
$$K(Q) = 0$$</p>

<p>If $Q &gt; 180$
$$L(Q) = 81$$
$$K(Q) = (Q - 180) / 2$$</p>

<p>Which you can get from thinking about what happens to the marginal cost of producing with labor just below and above $L=81$ (goes from below 4.5 to above 4.5) and comparing it with the the cost of producing a marginal unit with capital (which is always 4.5).</p>

<p>Why the split at $Q = 180 / L = 81 $?
 $$\frac{MP_{L}}{MP_{K}}=w/r \leftrightarrow  \frac{10L^{-1/2}}{2} = 5/9 \rightarrow L = 81$$
Note that $ Q(k=0,L=81) = 180$</p>

CC BY-SA 3.0
2015-03-11T13:14:49.980Z
2015-03-11T13:29:43.800Z
2015-03-11T13:29:43.800Z
BKay
<p><strong>A)</strong> <strong>If we care about not making losses,</strong></p>

<p>one can observe that  </p>

<p><strong>a)</strong> $Q(K=0, L&gt;0) &gt;0$  (i.e. we can obtain positive output without capital)  </p>

<p>and   </p>

<p><strong>b)</strong> Output is linear in $K$ and $MP_K = 2$, while $r=9$. So for each unit of capital we will employ,  we will obtain $2$ additional units of output, but we will have to pay $9$ units of output as capital reward: employing capital means subsidizing it here, given its price. So we should not employ capital at all, since in light of a), we can have positive output without it.</p>

<p>On the other hand if, say, $L=1$ then $Q = 20$ while $wL = 5$. So there is room for production to be carried using only labor, without subsidizing the production factor.</p>

<p>So we have $K^* = 0$. But then, </p>

<p>$$C^* =  wL^*,\;\; L^* = \frac {\bar Q^2}{400}$$</p>

<p>The cost function now has a non-interesting minimum at zero, and then it monotonically increases.</p>

<p>In order not to make losses, we need </p>

<p>$$ \bar Q - C^* \geq 0 \implies \bar Q - w\frac {\bar Q^2}{400} \geq 0 \implies \left(1-w\frac {\bar Q^*}{400}\right) \geq 0 $$</p>

<p>$$ \bar Q \leq \frac {400}{w}$$</p>

<p>So, <strong>if we care about not making losses</strong>, a) We do not employ capital and b) output level should not exceed, for $w=5$, $Q \leq 80$.</p>

<p><strong>B) If we don't care about making losses</strong> (say, this is a public utility that has to deliver a specific level of output irrespective of whether it will cost more than the revenue obtained from selling the output)</p>

<p>Let's see where the standard approach leads us.</p>

<p>$$\min_{K,L} C = rK + wL  \;\;\;s.t.\;\; g(K,L) = 2K + 20L^{1/2} = \bar Q$$</p>

<p>for any given level of output.</p>

<p>The Lagrangean is</p>

<p>$$ \Lambda = rK + wL + \lambda[\bar Q-2K - 20L^{1/2}]$$</p>

<p>and if we calculate the first-order conditions for a minimum we get</p>

<p>$$\partial \Lambda/\partial K = 0 \implies r - 2\lambda =0 \implies \lambda = r/2 \tag{1}$$</p>

<p>$$\partial \Lambda/\partial L = 0 \implies w - \lambda\cdot 10L^{-1/2} = 0 \tag {2}$$</p>

<p>which for $r=9, w=5$ leads to a <em>candidate</em> solution $\{K^*,L^*=81,\lambda^* = 4.5\}$</p>

<p>To determine what happens at the candidate solution we need to consider also second-order conditions for a minimum</p>

<p>$$\partial^2 \Lambda/\partial L^2 = 5\lambda L^{-3/2} $$</p>

<p>$$\partial^2 \Lambda/\partial K^2 = \partial^2 \Lambda/\partial L\partial K =0$$</p>

<p>Then the bordered Hessian matrix (=matrix of second derivatives of the Lagrangean bordered by the first derivatives of the constraint, and a zero in the upper left position) is</p>

<p>$$\bar H = \left [ \begin{matrix}
0 &amp; 2 &amp; 10L^{-1/2} \\
2 &amp; 0 &amp; 0 \\
10L^{-1/2} &amp; 0 &amp; 5\lambda L^{-3/2} \\
\end{matrix} \right]$$</p>

<p>For the candidate solution to be a minimum, we need the bordered principal minors (minor determinants) to all be <em>strictly negative</em> except for the first which is by construction zero), at least evaluated at the candidate solution. We have</p>

<p>$$|\bar H_2| = \left | \begin{matrix}
0 &amp; 2  \\
2 &amp; 0  \\
\end{matrix} \right| = -4 &lt;0$$</p>

<p>and </p>

<p>$$|\bar H_3| = \left | \begin{matrix}
0 &amp; 2 &amp; 10L^{-1/2} \\
2 &amp; 0 &amp; 0 \\
10L^{-1/2} &amp; 0 &amp; 5\lambda L^{-3/2} \\
\end{matrix} \right| = 0 - 2\cdot10\lambda\cdot L^{-3/2} + 0 &lt;0 $$</p>

<p>for $\lambda , L$ strictly positive. So we 're good... except we do not know the value for $K^*$. <strong>Are we free to choose any level of capital we want?</strong></p>

<p>Certainly not. Remember that the multiplier $\lambda ^*$  is <em>optimal</em> Marginal Cost (as a function of output). But here, <strong>assuming non-zero Capital is employed</strong>, <em>optimal marginal cost does not depend on the level of output, it is fixed at $\lambda ^* =4.5$</em>, due to the linear appearance of $K$ in the production function, and the assumed price-taking behavior of the organization.  </p>

<p>In other words: if we employ capital, we should have fixed $MC(Q) = 4.5$. This means that as long as we can carry production with lower marginal cost than that, we should not employ capital. But if there is a level of output produced solely by labor, at and after which, to continue producing using only labor would result in marginal cost higher than $4.5$, then, it becomes optimal to start employ capital (optimal strictly in the sense of cost-minimization).  </p>

<p>Under production solely using labor we have</p>

<p>$$C = \frac {w\bar Q^2}{400} \implies MC = \frac {\bar Q}{40}$$ </p>

<p>So for $MC \leq 4.5 \implies \bar Q \leq 180$
After $\bar Q =180$ if we employ additional labor to increase production our marginal cost will be higher than if we start employing capital. And this will continue, because the marginal product of labor will be falling, while the marginal product of capital is constant. So we conclude that up to $\bar Q =180$ we should carry production using only labor, and after that, we should stop employing any additional labor, and cover all resource needs by employing capital, the level of which will evidently be determined as given in @BKay 's answer.</p>

<p>The full mathematical treatment would require turning this into a Karush-Kuhn-Tucker framework, and include multipliers for the case of decision variables taking the value zero.</p>

CC BY-SA 3.0
2015-03-11T14:00:16.957Z
2015-03-11T15:29:19.850Z
2015-03-11T15:29:19.850Z
Alecos Papadopoulos
<p>So I have a production function $Q=2K + 20L^{1/2}$ and I suppose the wage is $w=5$ and the rental rate is $r=9$.  I want to find the long-run cost of production, which I know is constrained by $\frac{MP_{L}}{MP_{K}}=w/r$ and therefore</p>

<p>$$\frac{10L^{-1/2}}{2} = 5/9$$</p>

<p>Which implies $L=81$.  However, I don't see how to find my quantity of capital, $K$.  Any help would be appreciated.</p>

<production-function>
Determining cost from production function, wage, and rental rate
Addem
# Determining cost from production function, wage, and rental rate

(Asked by: username_0 on 2015-03-11)

So I have a production function $Q=2K + 20L^{1/2}$ and I suppose the wage is $w=5$ and the rental rate is $r=9$. I want to find the long-run cost of production, which I know is constrained by $\frac{MP\_{L}}{MP\_{K}}=w/r$ and therefore

$$\frac{10L^{-1/2}}{2} = 5/9$$

Which implies $L=81$. However, I don't see how to find my quantity of capital, $K$. Any help would be appreciated.

**Tags:** production-function


---

## 2 answers

### Answer by username_2 on 2015-03-11. Score: 3 | Accepted answer

**A)** **If we care about not making losses,**

one can observe that

**a)** $Q(K=0, L>0) >0$ (i.e. we can obtain positive output without capital)

and

**b)** Output is linear in $K$ and $MP\_K = 2$, while $r=9$. So for each unit of capital we will employ, we will obtain $2$ additional units of output, but we will have to pay $9$ units of output as capital reward: employing capital means subsidizing it here, given its price. So we should not employ capital at all, since in light of a), we can have positive output without it.

On the other hand if, say, $L=1$ then $Q = 20$ while $wL = 5$. So there is room for production to be carried using only labor, without subsidizing the production factor.

So we have $K^\* = 0$. But then,

$$C^\* = wL^\*,\;\; L^\* = \frac {\bar Q^2}{400}$$

The cost function now has a non-interesting minimum at zero, and then it monotonically increases.

In order not to make losses, we need

$$ \bar Q - C^\* \geq 0 \implies \bar Q - w\frac {\bar Q^2}{400} \geq 0 \implies \left(1-w\frac {\bar Q^\*}{400}\right) \geq 0 $$

$$ \bar Q \leq \frac {400}{w}$$

So, **if we care about not making losses**, a) We do not employ capital and b) output level should not exceed, for $w=5$, $Q \leq 80$.

**B) If we don't care about making losses** (say, this is a public utility that has to deliver a specific level of output irrespective of whether it will cost more than the revenue obtained from selling the output)

Let's see where the standard approach leads us.

$$\min\_{K,L} C = rK + wL \;\;\;s.t.\;\; g(K,L) = 2K + 20L^{1/2} = \bar Q$$

for any given level of output.

The Lagrangean is

$$ \Lambda = rK + wL + \lambda[\bar Q-2K - 20L^{1/2}]$$

and if we calculate the first-order conditions for a minimum we get

$$\partial \Lambda/\partial K = 0 \implies r - 2\lambda =0 \implies \lambda = r/2 \tag{1}$$

$$\partial \Lambda/\partial L = 0 \implies w - \lambda\cdot 10L^{-1/2} = 0 \tag {2}$$

which for $r=9, w=5$ leads to a *candidate* solution $\{K^\*,L^\*=81,\lambda^\* = 4.5\}$

To determine what happens at the candidate solution we need to consider also second-order conditions for a minimum

$$\partial^2 \Lambda/\partial L^2 = 5\lambda L^{-3/2} $$

$$\partial^2 \Lambda/\partial K^2 = \partial^2 \Lambda/\partial L\partial K =0$$

Then the bordered Hessian matrix (=matrix of second derivatives of the Lagrangean bordered by the first derivatives of the constraint, and a zero in the upper left position) is

$$\bar H = \left [ \begin{matrix}
0 & 2 & 10L^{-1/2} \\
2 & 0 & 0 \\
10L^{-1/2} & 0 & 5\lambda L^{-3/2} \\
\end{matrix} \right]$$

For the candidate solution to be a minimum, we need the bordered principal minors (minor determinants) to all be *strictly negative* except for the first which is by construction zero), at least evaluated at the candidate solution. We have

$$|\bar H\_2| = \left | \begin{matrix}
0 & 2 \\
2 & 0 \\
\end{matrix} \right| = -4 <0$$

and

$$|\bar H\_3| = \left | \begin{matrix}
0 & 2 & 10L^{-1/2} \\
2 & 0 & 0 \\
10L^{-1/2} & 0 & 5\lambda L^{-3/2} \\
\end{matrix} \right| = 0 - 2\cdot10\lambda\cdot L^{-3/2} + 0 <0 $$

for $\lambda , L$ strictly positive. So we 're good... except we do not know the value for $K^\*$. **Are we free to choose any level of capital we want?**

Certainly not. Remember that the multiplier $\lambda ^\*$ is *optimal* Marginal Cost (as a function of output). But here, **assuming non-zero Capital is employed**, *optimal marginal cost does not depend on the level of output, it is fixed at $\lambda ^\* =4.5$*, due to the linear appearance of $K$ in the production function, and the assumed price-taking behavior of the organization.

In other words: if we employ capital, we should have fixed $MC(Q) = 4.5$. This means that as long as we can carry production with lower marginal cost than that, we should not employ capital. But if there is a level of output produced solely by labor, at and after which, to continue producing using only labor would result in marginal cost higher than $4.5$, then, it becomes optimal to start employ capital (optimal strictly in the sense of cost-minimization).

Under production solely using labor we have

$$C = \frac {w\bar Q^2}{400} \implies MC = \frac {\bar Q}{40}$$

So for $MC \leq 4.5 \implies \bar Q \leq 180$
After $\bar Q =180$ if we employ additional labor to increase production our marginal cost will be higher than if we start employing capital. And this will continue, because the marginal product of labor will be falling, while the marginal product of capital is constant. So we conclude that up to $\bar Q =180$ we should carry production using only labor, and after that, we should stop employing any additional labor, and cover all resource needs by employing capital, the level of which will evidently be determined as given in @username_1 's answer.

The full mathematical treatment would require turning this into a Karush-Kuhn-Tucker framework, and include multipliers for the case of decision variables taking the value zero.


---

### Answer by username_1 on 2015-03-11. Score: 3

What's tricky is that the problem has a kink. Below a certain production value no capital is used. The demand functions for $k$ and $l$ are as follows:

If $Q \leq 180$
$$L(Q) = (Q/20)^2$$
$$K(Q) = 0$$

If $Q > 180$
$$L(Q) = 81$$
$$K(Q) = (Q - 180) / 2$$

Which you can get from thinking about what happens to the marginal cost of producing with labor just below and above $L=81$ (goes from below 4.5 to above 4.5) and comparing it with the the cost of producing a marginal unit with capital (which is always 4.5).

Why the split at $Q = 180 / L = 81 $?
$$\frac{MP\_{L}}{MP\_{K}}=w/r \leftrightarrow \frac{10L^{-1/2}}{2} = 5/9 \rightarrow L = 81$$
Note that $ Q(k=0,L=81) = 180$


---

