I am solving for the Leontief production function in the form of y= min {L^1/2 , K^1/2} , in the short run (fixed K) and in the long run respectively. I am asked to solve the production in terms of w, r the prices of labour and capital, and the sales price p.

I understand that for the Leontief function of the form:

y = min {z/a , z2/b} the optimal solution of inputs would be solved by letting z/a=z2/b then solving for z, but in this scenario, how would one express their answer in terms of w,r, and p?

  • $\begingroup$ What do you mean by ``solving the production function''. Are you looking at the profit maximisation problem or the cost minimisation problem. Also, could you maybe show what you have done so far in terms of trying to solve the question. $\endgroup$
    – tdm
    Commented Apr 8 at 5:37

2 Answers 2


Short Run Cost Minimisation problem:

\begin{eqnarray*} \min_{l\geq 0} & \ wl +rk \\ \text{s.t. } & y\leq \min(\sqrt{l},\sqrt{k}) \end{eqnarray*} where $y\geq 0$, $k\geq 0$, $w>0$, $r>0$ are given.

This problem has no solution if $y>\sqrt{k}$ because the set of $l$ satisfying the constraint is an empty set. Solving this problem for $y\leq \sqrt{k}$, we get the following short-run conditional labor demand function: \begin{eqnarray*} l^{sc}(w,r,k,y)=y^2\end{eqnarray*}

Therefore, short-run cost function is \begin{eqnarray*} c^{s}(w,r,k,y)=\begin{cases}wy^2+rk & \text{if } y\leq \sqrt{k}\\ \infty & \text{if }y>\sqrt{k}\end{cases} \end{eqnarray*}

where cost at $y>\sqrt{k}$ is $\infty$ suggests that it is impossible to reach an output of $y$ irrespective of how much labor is employed when capital is sufficiently low.

Short Run Profit Maximisation problem:

\begin{eqnarray*} \max_{y\geq 0} & \ py - c^s(w,r,k,y) \end{eqnarray*} where $p\geq 0$, $k\geq 0$, $w>0$, $r>0$ are given.

Solving this problem, we get the short-run supply as

\begin{eqnarray*} y^{ss}(p,k,w,r) =\min\left(\dfrac{p}{2w},\sqrt{k}\right)\end{eqnarray*}

Substituting this into the short-run conditional labor demand obtained earlier will provide us with the short-run labor demand as follows: \begin{eqnarray*} l^{sd}(p,k,w,r) =\min\left(\left(\dfrac{p}{2w}\right)^2,k\right)\end{eqnarray*}

Long Run Cost Minimisation problem: \begin{eqnarray*} \min_{l\geq 0,k\geq 0} & \ wl +rk \\ \text{s.t. } & y\leq \min(\sqrt{l},\sqrt{k}) \end{eqnarray*} where $y\geq 0$, $w>0$, $r>0$ are given.

Solving this problem we get the following long-run conditional labor and capital demand functions: \begin{eqnarray*} (l^{c},k^{c})(w,r,y)=(y^2,y^2)\end{eqnarray*}

Therefore, long-run cost function is \begin{eqnarray*} c(w,r,y)=wy^2+ry^2=(w+r)y^2 \end{eqnarray*}

Long Run Profit Maximisation problem:

\begin{eqnarray*} \max_{y\geq 0} & \ py - c(w,r,y) \end{eqnarray*} where $p\geq 0$, $w>0$, $r>0$ are given.

Solving this problem, we get the long-run supply as

\begin{eqnarray*} y^s(p,w,r) =\dfrac{p}{2(w+r)}\end{eqnarray*}

Substituting this into the long-run conditional labor demand obtained earlier will provide us with the long-run labor and capital demand as follows: \begin{eqnarray*} (l^d,k^d)(p,w,r) =\left(\left(\dfrac{p}{2(w+r)}\right)^2,\left(\dfrac{p}{2(w+r)}\right)^2\right)\end{eqnarray*}


I'll solve the Cost-Minimization and Profit-Maximization problems for the Leontief Production Function in the Long Run

Solution: Here we are given the production as, $$y=f(l,k)=min\{l^{0.5},k^{0.5}\}$$ i.e., we are dealing with a DRS Technology


Therefore, the Cost Minimization problem will be, $$min_{l,k}\hspace{3mm}(wl+r{k})$$ $$st. l \ge 0,y \le f(l,{k})=min\{l^{0.5},{ k}^{0.5}\}$$

The equilibrium will be when, $$y=l^{\frac{1}{2}}={k}^\frac{1}{2}$$ $$\Rightarrow y^2=l^\star={k}^\star$$

Now we will get the cost function as, $c(y;w,r)=wl+r{k}=(w+r)y^2$

Assuming that the firm is a price-taker, The Long Run Profit Maximization problem will now become,

$$max_y \hspace{4 mm} p(y)-(w+r)y^2$$ $$st. y\ge 0$$

Solving the above problem, we will get the Long-Run Supply Function of the Firm as , $$y^s(p,w,r)=\begin{cases}\frac{p}{2(w+r)}, & \text {if $p > 0$} \\ 0, &\text{otherwise}\end{cases}$$


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