Cost minimization problem

Question

The problem is as follows:

A firm uses $k$ units of capital and $l$ units of labor to produce $(k^{\alpha}l^{1-\alpha})^{1/\beta}$ units of output, where $\alpha$ and $\beta$ satisfying $0 < \alpha < 1$ and $\beta > 1$, are parameters.

If the firm obtains these inputs at the competitive rental rate $r$ for capital and the competitive wage rate $w$ for labor, and seeks to minimize the cost of producing at least $y$ units of output, it solves the problem

$$\min rk + wl \text{ subject to } (k^{\alpha}l^{1-\alpha})^{1/\beta}\geq y$$

where, in this cost-minimization problem, the output requirement y is another parameter.

(1) Using Kuhn-Tucker theorem, show that the minimum cost function $c(r,w,y)$ is

$$c(r,w,y) = \frac{1}{\alpha^{\alpha}\times(1-\alpha)^{(1-\alpha)}} \times r^{\alpha}w^{(1-\alpha)}y^\beta $$

(2) Using the Envelop theorem, and the equation (1) derive the optimal solutions $k*(r,w,y)$ and $l*(r,w,y)$.

This is my homework problem. Humble request to anyone for the solution. Thanks.

Answer 1

Set up the Lagrangian function:

$$\min_{k,l} rk+wl-\lambda\left((k^{\alpha}l^{1-\alpha})^{1/\beta}-y\right)$$

Note that the function $(k^{\alpha}l^{1-\alpha})^{1/\beta}$ is quasi-concave and homogeneous and of degree less than 1 (since $\beta>1$). A quasi-concave function of degree between 0 and 1 is concave. The negative of a concave function is convex ($\lambda$ is positive). Further, the two linear functions $rk$ and $wl$ are convex. The sum of convex functions is convex. Thus, the entire Lagrangian function is convex. The first order conditions will be sufficient for a minimum. It is also noted that the production function has infinite marginal product for either capital or labor when one is set to $0$. Thus, we will not have to worry about corner solutions where only one of the two inputs is used. Further, prices are strictly positive and so no more than y will be produced. The first order conditions are:

$$r=\lambda\frac{1}{\beta}\left(k^{\alpha}l^{\left(1-\alpha\right)}\right)^{\frac{1}{\beta}-1}\alpha\left(\frac{l}{k}\right)^{1-\alpha}$$

$$w=\lambda\frac{1}{\beta}\left(k^{\alpha}l^{\left(1-\alpha\right)}\right)^{\frac{1}{\beta}-1}\left(1-\alpha\right)\left(\frac{k}{l}\right)^{\alpha}$$

$$(k^{\alpha}l^{1-\alpha})^{1/\beta}=y$$

Solving these gives:

$$\frac{r}{\frac{1}{\beta}\left(k^{\alpha}l^{\left(1-\alpha\right)}\right)^{\frac{1}{\beta}-1}\alpha\left(\frac{l}{k}\right)^{1-\alpha}}=\frac{w}{\frac{1}{\beta}\left(k^{\alpha}l^{\left(1-\alpha\right)}\right)^{\frac{1}{\beta}-1}\left(1-\alpha\right)\left(\frac{k}{l}\right)^{\alpha}}$$

$$\frac{\left(1-\alpha\right)\left(\frac{k}{l}\right)^{\alpha}}{\alpha\left(\frac{l}{k}\right)^{1-\alpha}}=\frac{w}{r}$$

Gives one key relationship:

$$\frac{k}{l}=\frac{\alpha}{1-\alpha}\frac{w}{r}$$

Now, we use this along with the production constraint to solve for the inputs:

$$(k^{\alpha}l^{1-\alpha})=y^{\beta}$$

$$\left(\frac{\alpha}{1-\alpha}\frac{w}{r}\right)^{\alpha}l=y^{\beta}$$

$$l^{*}=\left(\frac{1-\alpha}{\alpha}\frac{r}{w}\right)^{\alpha}y^{\beta}$$

$$k^{*}=\left(\frac{\alpha}{1-\alpha}\frac{w}{r}\right)^{1-\alpha}y^{\beta}$$

Plugging these back into the cost function gives the desired result:

$$y^{\beta}w^{1-\alpha}r^{\alpha}\left[\left(\frac{\alpha}{1-\alpha}\right)^{1-\alpha}+\left(\frac{1-\alpha}{\alpha}\right)^{\alpha}\right]$$

$$y^{\beta}w^{1-\alpha}r^{\alpha}\left[\left(1+\left(\frac{\alpha}{1-\alpha}\right)\right)\left(\frac{1-\alpha}{\alpha}\right)^{\alpha}\right]$$

$$y^{\beta}w^{1-\alpha}r^{\alpha}\left[\left(\frac{1}{1-\alpha}\right)\left(\frac{1-\alpha}{\alpha}\right)^{\alpha}\right]$$

$$y^{\beta}w^{1-\alpha}r^{\alpha}\frac{1}{\alpha^{\alpha}\left(1-\alpha\right)^{1-\alpha}}$$

By the envelope theorem, the derivative of an optimized function with respect to one of its parameters can be taken treating the choice variables as fixed. At the optimum, a very small change in these choice variables has no effect on the optimized value. This means, when we take the derivative of the optimized cost function with respect to $r$ and $w$ we will get back $k^*$ and $l^*$ respectively. I will leave this to you to check.