<p>In this thread, two problems are discussed:</p>
<p><strong>Problem 1</strong>:
<span class="math-container">\begin{eqnarray*} \max_{l, k} &amp; \ 16 l^{\frac{1}{4}}k^{\frac{3}{4}} \\ \text{s.t. } &amp; 50l + 100k \leq 500,000 \\ \text{and } &amp; l\geq 0, k\geq 0\end{eqnarray*}</span>
Solution to this problem satisfy:</p>
<p><span class="math-container">$\text{MRTS} = \dfrac{\text{MP}_L}{\text{MP}_K}=\dfrac{k}{3l} = \dfrac{50}{100}$</span>
and
<span class="math-container">$50l+100k = 500,000$</span></p>
<p>Solving the system, we get
<span class="math-container">$(l, k) = (2500,3750)$</span></p>
<p><strong>Problem 2</strong>:
<span class="math-container">\begin{eqnarray*} \max_{l, k} &amp; \ p(16 l^{\frac{1}{4}}k^{\frac{3}{4}}) -50l-100k \\ \text{s.t. } &amp; 50l + 100k \leq 500,000 \\ \text{and } &amp; l\geq 0, k\geq 0\end{eqnarray*}</span></p>
<p>To solve the above problem, we can first solve the cost-minimization problem of the firm.</p>
<p><span class="math-container">\begin{eqnarray*} \min_{l, k} &amp; 50l+100k \\ \text{s.t. } &amp; 16 l^{\frac{1}{4}}k^{\frac{3}{4}} \geq q \\ \text{and } &amp; l\geq 0, k\geq 0\end{eqnarray*}</span></p>
<p>Solution to this problem satisfy:</p>
<p><span class="math-container">$\text{MRTS} = \dfrac{\text{MP}_L}{\text{MP}_K}=\dfrac{k}{3l} = \dfrac{50}{100}$</span>
and
<span class="math-container">$16 l^{\frac{1}{4}}k^{\frac{3}{4}} = q$</span></p>
<p>Solving the system, we get the conditional labor and capital demand as:
<span class="math-container">$(l^c, k^c)(q) = \left(\left(\frac{2}{3}\right)^\frac{3}{4} \frac{q}{16}, \left(\frac{3}{2}\right)^\frac{1}{4} \frac{q}{16}\right)$</span></p>
<p>and the cost function is
<span class="math-container">$C(q) = \dfrac{50 \left(\frac{2}{3}\right)^\frac{3}{4} + 100 \left(\frac{3}{2}\right)^\frac{1}{4}}{16}q$</span></p>
<p>Let's define <span class="math-container">$c = \dfrac{50 \left(\frac{2}{3}\right)^\frac{3}{4} + 100 \left(\frac{3}{2}\right)^\frac{1}{4}}{16}$</span></p>
<p>Therefore, <span class="math-container">$C(q) = cq$</span>.</p>
<p>Now we can re-write the above profit maximization problem as:
<span class="math-container">\begin{eqnarray*} \max_{q} &amp; \ pq -cq \\ \text{s.t. } &amp; cq \leq 500,000 \\ \text{and } &amp; q \geq 0\end{eqnarray*}</span></p>
<p>Solving this problem, gives the supply:
<span class="math-container">\begin{eqnarray*} q^s(p) \in \begin{cases} \left\{\frac{500,000}{c}\right\} &amp; \text{if } p &gt; c \\ \left[0,\frac{500,000}{c}\right] &amp; \text{if } p = c \\ \{0\} &amp; \text{if } p &lt; c \end{cases} \end{eqnarray*}</span></p>
<p>Substituting the supply in conditional labor demand gives the labor demand as:
<span class="math-container">\begin{eqnarray*} l^d(p) \in \begin{cases} \left\{2500\right\} &amp; \text{if } p &gt; c \\ \left[0,2500\right] &amp; \text{if } p = c \\ \{0\} &amp; \text{if } p &lt; c \end{cases} \end{eqnarray*}</span>
and capital demand is <span class="math-container">$k^d(p)=\frac{3}{2}l^d(p)$</span></p>

CC BY-SA 4.0
2022-06-28T17:35:29.507Z
2022-06-28T17:35:29.507Z
Amit
<p>Based on Macosso's comment, I'll present another way for finding the local maximum of the function under consideration such that the budget constraint is met. Instead of Lagrange's multiplier, we investigate the parametric system of functions and reduce the original maximization problem to a simple maximization of polynomial values.</p>
<p>Our aim is to maximize the profit function
<span class="math-container">$$F(K,L)
 :=P\cdot16\cdot L^{\frac14}\cdot K^{\frac34}-(50L+100K)$$</span>
with the budget condition <span class="math-container">$50L+100K\le5\cdot 10^5$</span> with <span class="math-container">$K,L\ge 0$</span>. This condition can be rewritten as <span class="math-container">$50L+100K=c$</span>, where <span class="math-container">$c$</span> is arbitrary fixed parameter with <span class="math-container">$0\le c\le5\cdot10^5$</span>. Consequently, this yields <span class="math-container">$L+2K=c/50$</span>, i.e., <span class="math-container">$L=c/50-2K$</span>. This enables us to reduce the number of the variables in the function <span class="math-container">$F$</span> defined above. We obtain
<span class="math-container">$$
F(K,L)\biggl |_{L=\frac c{50}-2K}
 =\underbrace{16\cdot P\cdot\sqrt[4]{\left(\frac c{50}-2K\right)\cdot K^3}-c}_{\varphi(K)}.
$$</span>
Remeber that <span class="math-container">$c$</span> is a fixed parameter. Hence, to find the local extrema of <span class="math-container">$\varphi(K)$</span>, one can neglect the constant additive parameter <span class="math-container">$c$</span> and investigate the function
<span class="math-container">$$
16\cdot P\cdot\sqrt[4]{\left(\frac c{50}-2K\right)\cdot K^3}.
$$</span>
Nevertheless, taking into account that <span class="math-container">$\sqrt[4]x$</span> is strictly increasing, we deduce that it preserves the monotonicity and, hence, the local extrema and their type (min., max.). Therefore, the functions
<span class="math-container">$$
f(K):=\left(\frac c{50}-2K\right)\cdot K^3
$$</span>
and <span class="math-container">$\varphi(K)$</span> attain their local extrema at the same <span class="math-container">$K$</span>-coordinate(s). Expanding the last polynomial expression implies
<span class="math-container">$$
\begin{align*}
f(K)
 &amp;=\frac c{50}\cdot K^3-2K^4,\\[4mm]
f'(K)
 &amp;=\frac{3c}{50}\cdot K^2-8K^3
  =K^2\cdot\underbrace{\left(\frac{3c}{50}-8K\right)}_{\text{zero point: $K=\frac{3c}{400}$}}.
\end{align*}
$$</span>
Since the derivative <span class="math-container">$f'(K)$</span> is positive in the left neighbourhood of <span class="math-container">$K=3c/400$</span> and negative in the right one, it follows that <span class="math-container">$f(K)$</span> attains its local maximum at this point.</p>
<p>Let us now evaluate the value of this maximum:
<span class="math-container">$$
\begin{align*}
\varphi\left(\frac{3c}{400}\right)
 &amp;=16\cdot P\cdot\sqrt[4]{\left(\frac c{50}-\frac{3c}{200}\right )\cdot\left(\frac{3c}{400}\right)^3}-c\\[1em]
 &amp;=P\cdot c\cdot\frac{\sqrt[4]{54}}{25}-c\\[1em]
 &amp;=c\cdot\left(P\cdot\frac{\sqrt[4]{54}}{25}-1\right).
\end{align*}
$$</span></p>
<p>Since we are searching for
<span class="math-container">$$
\max_{\substack{K,L\ge 0\\50L+100K\le 5\cdot 10^5}}F(K,L),
$$</span>
it finally suffices to find the maximum of <span class="math-container">$\varphi(3c/400)$</span> with <span class="math-container">$0\le c\le 5\cdot 10^5$</span>. We easily obtain that
<span class="math-container">$$
\begin{align*}
\max_{\substack{K,L\ge 0\\50L+100K\le 5\cdot 10^5}}F(K,L)
 &amp;=\max_{0\le c\le 5\cdot10^5}\varphi\left(\frac{3c}{400}\right)\\[1em]
 &amp;=\max_{0\le c\le 5\cdot10^5}c\cdot\left(P\cdot\frac{\sqrt[4]{54}}{25}-1\right)\\[1.5em]
 &amp;=\begin{cases}
   5\cdot 10^5\cdot\left(P\cdot\frac{\sqrt[4]{54}}{25}-1\right)&amp; \text{for $P&gt;\frac{25}{\sqrt[4]{54}}$},\\[1em]
   0&amp; \text{for $0&lt;P\le\frac{25}{\sqrt[4]{54}}$}.
  \end{cases}
\end{align*}
$$</span></p>
<p>Of course, to localise the extreme (optimum), we can write (regarding the first case <span class="math-container">$P&gt;25/\sqrt[4]{54}$</span>)
<span class="math-container">$$
\begin{align*}
K_0&amp;=\frac{3c}{400}\Biggl|_{c=5\cdot 10^5}=3750,\\[1em]
L_0&amp;=\frac c{50}-2K_0\Biggl|_{c=5\cdot 10^5}=2500.
\end{align*}
$$</span></p>
<p>Remark: I think it was not obvious to consider the value <span class="math-container">$c=5\cdot 10^5$</span> right from the beginning.</p>

CC BY-SA 4.0
2021-12-01T23:20:28.940Z
2021-12-01T23:55:44.520Z
2021-12-01T23:55:44.520Z
Marian G.
<p>If your aim is to maximize the production then your approach is correct.</p>
<p>But if the aim is to find the optimal number of units of labor, then you should solve it as a profit maximization problem with a budget constraint.</p>
<p>Then the problem should be as following</p>
<p>Maximize the profit such that the total cost spent on production does not exceed the budget
<span class="math-container">$$ Max_{L,K}\ Profit(K,L)=P(16L^\frac{1}{4}K^\frac{3}{4}) - 50L - 100K $$</span>
<span class="math-container">$$st. 50L + 100K \le 500000$$</span></p>
<p>Then the Lagrange function would be:
<span class="math-container">$$ \mathcal{L}(K,L,\lambda) = (P(16L^\frac{1}{4}K^\frac{3}{4}) - 50L - 100K) + \lambda(500,000-50L-100K)$$</span></p>
<p>FOC:</p>
<p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial L}=4PL^\frac{-3}{4}K^\frac{3}{4} - 50-50\lambda =0$</span></p>
<p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial K}=12PL^\frac{1}{4}K^\frac{-1}{4} - 100-100\lambda =0$</span></p>
<p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial \lambda}=500,000 - 50L - 100K =0$</span></p>
<p>Solving the system of FOC equations will yield <span class="math-container">$L = 2500, K = 3750$</span>
I used Wolfram to get the solution, but you can solve that using the approach you used in the question.</p>
<p>Wolfram code:</p>
<p>Z = Lambda</p>
<pre><code>Solve[{4P (L^(-3/4))(K^(3/4)) - 50 - 50Z==0, 12P (L^(1/4))(K^(-1/4)) - 100 - 100Z==0, 500000-50L-100K==0},{L,K,Z}]
</code></pre>

CC BY-SA 4.0
2021-11-28T21:54:03.783Z
@Macosso: I have two points. First, "total cost spent on production does not exceed the budget" means $50L+100K\le 500\,000$, i.e., the opposite inequality sign. The second one is a technical remark. In fact, there is no need to use the method of Lagrange multiplier. Modifying the budget condition to $L=10^4-2K$ enables one to reduce the number of variables in $F:=16P(LK^3)^{1/4}-50L-100K$. Then $F$ becomes $16P(10^4K^3-2K^4)^{1/4}$. Since $x^{1/4}$ is increasing, it suffices to maximize the polynomial $10^4K^3-2K^4$. Of course, this yields identical result as in your post.
Marian G.
CC BY-SA 4.0
2021-11-29T07:33:29.450Z
Yes, you are righr for point 1 and 2. 1. The constraint should should be $\le$ the sign has been updated. 2. You dont have to use the Lagrange methods when there are only 2 variables, you can just solve for for one variable in th budget constraint and plug it to the production function. And this reduces the amount of variables. But we tend to use more Lagrange method because it is more powerful, when we have multiple variables etc. In this specific case I used the Lagrange method because the question was asked with initial solutions thar involved Lagrange.
Macosso
CC BY-SA 4.0
2021-11-29T14:07:29.650Z
@MarianG. it would be great to add your point 2 as an additional answer, it would be very helpful for people looking for different ways of solving this kind of problems.
Macosso
CC BY-SA 4.0
2021-11-29T15:00:21.387Z
OK, I'll add it in the evening. Thanks for your feedback.
Marian G.
CC BY-SA 4.0
2021-12-01T23:35:18.893Z
@Macosso: In my contribution, I added the idea of finding the required local extreme. Also, a discussion about the parameter $P$ is presented.
Marian G.
CC BY-SA 4.0
2022-06-29T05:21:34.807Z
@Macosso The solution you obtained is independent of the price $P$ which cannot be correct. If $P$ is small enough, consider some value of $P$ that is close enough to $0$, in that case hiring 2500 units of labor and 3750 units of capital will lead to losses. A better choice in that case is to hire 0 units of labor and 0 units of capital since that would lead to 0 profits (which is better than loss).
Amit
CC BY-SA 4.0
2021-11-23T17:51:15.377Z
2021-11-29T07:05:20.587Z
2021-11-29T07:05:20.587Z
Macosso
<p>Given production function <span class="math-container">$f(L,K)=16L^\frac{1}{4}K^\frac{3}{4}$</span>, where each unit of labour costs £50 and each unit of capital costs £100 and you have a budget of £500,000. Find the number of units of labour to maximise production.</p>
<p>We were given this question in a maths class, so I am slightly unsure of my solution, as follows:</p>
<p>We have <span class="math-container">$Y = 16L^\frac{1}{4}K^\frac{3}{4} + \lambda(500,000-50L-100K)$</span></p>
<p><span class="math-container">$\frac{\partial Y}{\partial L}=4L^\frac{-3}{4}K^\frac{3}{4}-50\lambda =0$</span></p>
<p><span class="math-container">$\frac{\partial Y}{\partial K}=12L^\frac{1}{4}K^\frac{-1}{4}-100\lambda =0$</span></p>
<p><span class="math-container">$\frac{\partial Y}{\partial \lambda}=500,000 - 50L - 100K =0$</span></p>
<p>Then, <span class="math-container">$\frac{(\frac{\partial Y}{\partial L})}{(\frac{\partial Y}{\partial K})} = \frac{4L^\frac{-3}{4}K^\frac{3}{4}}{12L^\frac{1}{4}K^\frac{-1}{4}} = \frac{50 \lambda}{100 \lambda} $</span></p>
<p><span class="math-container">$\implies \frac{K}{3L}=\frac{1}{2} \implies k = \frac{3}{2}L$</span></p>
<p>Then inputting into <span class="math-container">$\frac{\partial y}{\partial \lambda}, 500,000 = 50L + 100(\frac{3}{2}L) = 200L$</span></p>
<p><span class="math-container">$\implies L = 2,500$</span></p>
<p>Therefore, at a budget of £500,000, 2,500 units of labour maximises production?</p>

<microeconomics><labor-economics><production-function><cobb-douglas><capital>
Cobb-Douglas Production Function - Finding units of labour to maximise production
James Burton
# Cobb-Douglas Production Function - Finding units of labour to maximise production

(Asked by: username_0 on 2021-11-23)

Given production function $f(L,K)=16L^\frac{1}{4}K^\frac{3}{4}$, where each unit of labour costs £50 and each unit of capital costs £100 and you have a budget of £500,000. Find the number of units of labour to maximise production.

We were given this question in a maths class, so I am slightly unsure of my solution, as follows:

We have $Y = 16L^\frac{1}{4}K^\frac{3}{4} + \lambda(500,000-50L-100K)$

$\frac{\partial Y}{\partial L}=4L^\frac{-3}{4}K^\frac{3}{4}-50\lambda =0$

$\frac{\partial Y}{\partial K}=12L^\frac{1}{4}K^\frac{-1}{4}-100\lambda =0$

$\frac{\partial Y}{\partial \lambda}=500,000 - 50L - 100K =0$

Then, $\frac{(\frac{\partial Y}{\partial L})}{(\frac{\partial Y}{\partial K})} = \frac{4L^\frac{-3}{4}K^\frac{3}{4}}{12L^\frac{1}{4}K^\frac{-1}{4}} = \frac{50 \lambda}{100 \lambda} $

$\implies \frac{K}{3L}=\frac{1}{2} \implies k = \frac{3}{2}L$

Then inputting into $\frac{\partial y}{\partial \lambda}, 500,000 = 50L + 100(\frac{3}{2}L) = 200L$

$\implies L = 2,500$

Therefore, at a budget of £500,000, 2,500 units of labour maximises production?

**Tags:** microeconomics, labor-economics, production-function, cobb-douglas, capital


---

## 3 answers

### Answer by username_3 on 2021-11-23. Score: 5 | Accepted answer

If your aim is to maximize the production then your approach is correct.

But if the aim is to find the optimal number of units of labor, then you should solve it as a profit maximization problem with a budget constraint.

Then the problem should be as following

Maximize the profit such that the total cost spent on production does not exceed the budget
$$ Max\_{L,K}\ Profit(K,L)=P(16L^\frac{1}{4}K^\frac{3}{4}) - 50L - 100K $$
$$st. 50L + 100K \le 500000$$

Then the Lagrange function would be:
$$ \mathcal{L}(K,L,\lambda) = (P(16L^\frac{1}{4}K^\frac{3}{4}) - 50L - 100K) + \lambda(500,000-50L-100K)$$

FOC:

$\frac{\partial \mathcal{L}}{\partial L}=4PL^\frac{-3}{4}K^\frac{3}{4} - 50-50\lambda =0$

$\frac{\partial \mathcal{L}}{\partial K}=12PL^\frac{1}{4}K^\frac{-1}{4} - 100-100\lambda =0$

$\frac{\partial \mathcal{L}}{\partial \lambda}=500,000 - 50L - 100K =0$

Solving the system of FOC equations will yield $L = 2500, K = 3750$
I used Wolfram to get the solution, but you can solve that using the approach you used in the question.

Wolfram code:

Z = Lambda

```
Solve[{4P (L^(-3/4))(K^(3/4)) - 50 - 50Z==0, 12P (L^(1/4))(K^(-1/4)) - 100 - 100Z==0, 500000-50L-100K==0},{L,K,Z}]

```

> **Comments:**
> **username_2**: @username_3: I have two points. First, "total cost spent on production does not exceed the budget" means $50L+100K\le 500\,000$, i.e., the opposite inequality sign. The second one is a technical remark. In fact, there is no need to use the method of Lagrange multiplier. Modifying the budget condition to $L=10^4-2K$ enables one to reduce the number of variables in $F:=16P(LK^3)^{1/4}-50L-100K$. Then $F$ becomes $16P(10^4K^3-2K^4)^{1/4}$. Since $x^{1/4}$ is increasing, it suffices to maximize the polynomial $10^4K^3-2K^4$. Of course, this yields identical result as in your post.  
> **username_3**: Yes, you are righr for point 1 and 2. 1. The constraint should should be $\le$ the sign has been updated. 2. You dont have to use the Lagrange methods when there are only 2 variables, you can just solve for for one variable in th budget constraint and plug it to the production function. And this reduces the amount of variables. But we tend to use more Lagrange method because it is more powerful, when we have multiple variables etc. In this specific case I used the Lagrange method because the question was asked with initial solutions thar involved Lagrange.  
> **username_3**: @MarianG. it would be great to add your point 2 as an additional answer, it would be very helpful for people looking for different ways of solving this kind of problems.  
> **username_2**: OK, I'll add it in the evening. Thanks for your feedback.  
> **username_2**: @username_3: In my contribution, I added the idea of finding the required local extreme. Also, a discussion about the parameter $P$ is presented.  
> **username_1**: @username_3 The solution you obtained is independent of the price $P$ which cannot be correct. If $P$ is small enough, consider some value of $P$ that is close enough to $0$, in that case hiring 2500 units of labor and 3750 units of capital will lead to losses. A better choice in that case is to hire 0 units of labor and 0 units of capital since that would lead to 0 profits (which is better than loss).  

---

### Answer by username_1 on 2022-06-28. Score: 3

In this thread, two problems are discussed:

**Problem 1**:
\begin{eqnarray\*} \max\_{l, k} & \ 16 l^{\frac{1}{4}}k^{\frac{3}{4}} \\ \text{s.t. } & 50l + 100k \leq 500,000 \\ \text{and } & l\geq 0, k\geq 0\end{eqnarray\*}
Solution to this problem satisfy:

$\text{MRTS} = \dfrac{\text{MP}\_L}{\text{MP}\_K}=\dfrac{k}{3l} = \dfrac{50}{100}$
and
$50l+100k = 500,000$

Solving the system, we get
$(l, k) = (2500,3750)$

**Problem 2**:
\begin{eqnarray\*} \max\_{l, k} & \ p(16 l^{\frac{1}{4}}k^{\frac{3}{4}}) -50l-100k \\ \text{s.t. } & 50l + 100k \leq 500,000 \\ \text{and } & l\geq 0, k\geq 0\end{eqnarray\*}

To solve the above problem, we can first solve the cost-minimization problem of the firm.

\begin{eqnarray\*} \min\_{l, k} & 50l+100k \\ \text{s.t. } & 16 l^{\frac{1}{4}}k^{\frac{3}{4}} \geq q \\ \text{and } & l\geq 0, k\geq 0\end{eqnarray\*}

Solution to this problem satisfy:

$\text{MRTS} = \dfrac{\text{MP}\_L}{\text{MP}\_K}=\dfrac{k}{3l} = \dfrac{50}{100}$
and
$16 l^{\frac{1}{4}}k^{\frac{3}{4}} = q$

Solving the system, we get the conditional labor and capital demand as:
$(l^c, k^c)(q) = \left(\left(\frac{2}{3}\right)^\frac{3}{4} \frac{q}{16}, \left(\frac{3}{2}\right)^\frac{1}{4} \frac{q}{16}\right)$

and the cost function is
$C(q) = \dfrac{50 \left(\frac{2}{3}\right)^\frac{3}{4} + 100 \left(\frac{3}{2}\right)^\frac{1}{4}}{16}q$

Let's define $c = \dfrac{50 \left(\frac{2}{3}\right)^\frac{3}{4} + 100 \left(\frac{3}{2}\right)^\frac{1}{4}}{16}$

Therefore, $C(q) = cq$.

Now we can re-write the above profit maximization problem as:
\begin{eqnarray\*} \max\_{q} & \ pq -cq \\ \text{s.t. } & cq \leq 500,000 \\ \text{and } & q \geq 0\end{eqnarray\*}

Solving this problem, gives the supply:
\begin{eqnarray\*} q^s(p) \in \begin{cases} \left\{\frac{500,000}{c}\right\} & \text{if } p > c \\ \left[0,\frac{500,000}{c}\right] & \text{if } p = c \\ \{0\} & \text{if } p < c \end{cases} \end{eqnarray\*}

Substituting the supply in conditional labor demand gives the labor demand as:
\begin{eqnarray\*} l^d(p) \in \begin{cases} \left\{2500\right\} & \text{if } p > c \\ \left[0,2500\right] & \text{if } p = c \\ \{0\} & \text{if } p < c \end{cases} \end{eqnarray\*}
and capital demand is $k^d(p)=\frac{3}{2}l^d(p)$


---

### Answer by username_2 on 2021-12-01. Score: 1

Based on username_3's comment, I'll present another way for finding the local maximum of the function under consideration such that the budget constraint is met. Instead of Lagrange's multiplier, we investigate the parametric system of functions and reduce the original maximization problem to a simple maximization of polynomial values.

Our aim is to maximize the profit function
$$F(K,L)
:=P\cdot16\cdot L^{\frac14}\cdot K^{\frac34}-(50L+100K)$$
with the budget condition $50L+100K\le5\cdot 10^5$ with $K,L\ge 0$. This condition can be rewritten as $50L+100K=c$, where $c$ is arbitrary fixed parameter with $0\le c\le5\cdot10^5$. Consequently, this yields $L+2K=c/50$, i.e., $L=c/50-2K$. This enables us to reduce the number of the variables in the function $F$ defined above. We obtain
$$
F(K,L)\biggl |\_{L=\frac c{50}-2K}
=\underbrace{16\cdot P\cdot\sqrt[4]{\left(\frac c{50}-2K\right)\cdot K^3}-c}\_{\varphi(K)}.
$$
Remeber that $c$ is a fixed parameter. Hence, to find the local extrema of $\varphi(K)$, one can neglect the constant additive parameter $c$ and investigate the function
$$
16\cdot P\cdot\sqrt[4]{\left(\frac c{50}-2K\right)\cdot K^3}.
$$
Nevertheless, taking into account that $\sqrt[4]x$ is strictly increasing, we deduce that it preserves the monotonicity and, hence, the local extrema and their type (min., max.). Therefore, the functions
$$
f(K):=\left(\frac c{50}-2K\right)\cdot K^3
$$
and $\varphi(K)$ attain their local extrema at the same $K$-coordinate(s). Expanding the last polynomial expression implies
$$
\begin{align\*}
f(K)
&=\frac c{50}\cdot K^3-2K^4,\\[4mm]
f'(K)
&=\frac{3c}{50}\cdot K^2-8K^3
=K^2\cdot\underbrace{\left(\frac{3c}{50}-8K\right)}\_{\text{zero point: $K=\frac{3c}{400}$}}.
\end{align\*}
$$
Since the derivative $f'(K)$ is positive in the left neighbourhood of $K=3c/400$ and negative in the right one, it follows that $f(K)$ attains its local maximum at this point.

Let us now evaluate the value of this maximum:
$$
\begin{align\*}
\varphi\left(\frac{3c}{400}\right)
&=16\cdot P\cdot\sqrt[4]{\left(\frac c{50}-\frac{3c}{200}\right )\cdot\left(\frac{3c}{400}\right)^3}-c\\[1em]
&=P\cdot c\cdot\frac{\sqrt[4]{54}}{25}-c\\[1em]
&=c\cdot\left(P\cdot\frac{\sqrt[4]{54}}{25}-1\right).
\end{align\*}
$$

Since we are searching for
$$
\max\_{\substack{K,L\ge 0\\50L+100K\le 5\cdot 10^5}}F(K,L),
$$
it finally suffices to find the maximum of $\varphi(3c/400)$ with $0\le c\le 5\cdot 10^5$. We easily obtain that
$$
\begin{align\*}
\max\_{\substack{K,L\ge 0\\50L+100K\le 5\cdot 10^5}}F(K,L)
&=\max\_{0\le c\le 5\cdot10^5}\varphi\left(\frac{3c}{400}\right)\\[1em]
&=\max\_{0\le c\le 5\cdot10^5}c\cdot\left(P\cdot\frac{\sqrt[4]{54}}{25}-1\right)\\[1.5em]
&=\begin{cases}
5\cdot 10^5\cdot\left(P\cdot\frac{\sqrt[4]{54}}{25}-1\right)& \text{for $P>\frac{25}{\sqrt[4]{54}}$},\\[1em]
0& \text{for $0<P\le\frac{25}{\sqrt[4]{54}}$}.
\end{cases}
\end{align\*}
$$

Of course, to localise the extreme (optimum), we can write (regarding the first case $P>25/\sqrt[4]{54}$)
$$
\begin{align\*}
K\_0&=\frac{3c}{400}\Biggl|\_{c=5\cdot 10^5}=3750,\\[1em]
L\_0&=\frac c{50}-2K\_0\Biggl|\_{c=5\cdot 10^5}=2500.
\end{align\*}
$$

Remark: I think it was not obvious to consider the value $c=5\cdot 10^5$ right from the beginning.


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