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I was doing my homework and got really confused about how to approach the optimal levels of inputs when there are three variables. My current understanding is that the problem is to solve the optimization problem

$$\min_{H, L, K}\;{sH + wL +rK}$$ subject to $$q = \min\{H,L\} + \min\{H,K\}$$

However, I don't know how to determine the inputs of H, L, and K. Below is the problem:

A monopolist can hire high skill labor $H$, low skill labor $L$, and robots $K$. The per-unit price of these inputs are $s$, $w$, and $r$. High skill labor is more expensive: $s > w$. The production function is $f(H, L, K) = \min\{H, L\} + \min\{H, K\}$. The demand curve that the monopolist faces is $D(p) = A-p$, where $A > w + r + s$. Suppose that robots are cheap: $r < w + s$.

Essentially, we're asked to find the optimal levels (yields lowest costs) of input in terms of q for H,L,K

I initially thought since $r<w+s$, and $w<s$, we can deduce that $r<s+s$, and for this reason, the production function would give either $q = H+K$, $q = L+H$, or $q = L+K$ (making $H+H$ the only impossible combination). However, I'm super unsure if this is the right thought process, and I don't know how to proceed from here.

Any help is greatly appreciated.

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I`d propose you to follow these steps:

  1. Set up the minimization cost problem (i.e. for a given output quantity $y$ minimize costs): \begin{align} \min_{H,L,K}& \quad sH + wL + rK \tag{1} \label{1}\\ \text{such that} &\quad \min\{H,L\} + \min\{H, K\}\geq y \tag{2} \label{2} \end{align}
  2. In principle you have 3 cases, depending on price of factors $(s,w,r)$.

    1) When $r>s+w$, then buying a unit of $L$ and one of $H$ is going to cost you strictly less than buying a unit ok $K$. Consequently the monopolist opts for producing only using $H$ and $L$ ($K=0$). Thus $H = L = y$

2) The second case happens for $s>r+w$ which is, anyway, ruled out by the assupmption $w>s$ (if this was not the case, then producing only using $H$ and $K$ would have been optimal, i.e $H = K = y$ and $L = 0$).

3) Finally, when none of the above occurs, you have $L = H$ and $K = H$ because of the fixed proportions production function. Then the constraint \eqref{2} becomes $H + H \geq y$ and the inequality sign become equality. The equality sign comes from the fact that producing more than $y$ (the LHS of \ref{2} being strictly larger than the RHS) is just overshooting and inflates production costs for a given $y$. $L = H$ and $K = H$ comes from the Leontief production set-up of the problem: inside the $\min\{...,...\}$ operator is never convenient to exploit more one of the two factors (because by decreaing the most used factor one still gets the same amount of output but a lower cost). Therefore, each of the two $\min\{...,...\}$ can just be seen as one of its inner arguments, namely, for convenience, $H$. So you finally get $H+H = y$ and, from there: $H = y/2$ and $L = K = y/2$

  1. Substitute the optimal values ${\big(H^*\small{(s,w,r)},L^*\small{(s,w,r)},K^*\small{(s,w,r)}\big)}$ in \eqref{1} so to get an expression only in $(s,w,r)$. This expression is the cost function $C(s,w,r,y)$
  2. Now solve the monopolist's pricing problem for given factor prices: \begin{equation} \max_p \quad p\cdot D(p) - C\big(s,w,r,D(p)\big) \end{equation} and find the profit maximizing price, $p^*$ for output level $y^* = D(p^*)$

I hope this helps, comment below if you need any more hint.

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  • $\begingroup$ I understand how to solve the problem but could you explain a little more about how you arrived at point 2? Thank you so much... $\endgroup$ – user24692 Oct 28 at 4:11
  • $\begingroup$ I`ve just edit my answer, hopes it clarifies enough. $\endgroup$ – GabMac Oct 28 at 8:22
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@GabMac, I don't think L=H, K=H will always hold given this function. I think L=H, K=H holds because of the inequality r < w + s. If r > w + s, the producer will always produce in H,L only and K=0 even when the production function remains fixed proportions. The answer is correct but I think only the reasoning for modifying the production function is wrong. Thus, L=H,K=H because of the given inequality r < w + s.

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  • $\begingroup$ You're right, I`ve edited my answer accordingly $\endgroup$ – GabMac Oct 28 at 22:14

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