Missing Non-Negativity Constraint on Lagrangian

Question

We have the constrained maximisation problem:

A perfectly competitive firm produces one output with two inputs, capital $(k)$ and labour $(l)$. The rental cost of capital is equal to $r >0$ and the wage rate is equal to $w>0$. The production function is $f(k, l)=(k + 1)^α l^{1-α}$ with $0 <α<1$. The firm wishes to minimise total costs while achieving a given level of production $y>0$.

Now for the sake of this course we construct this as a maximisation, problem I am particularly interested in the case where $k=0$ and $l>0$. While they do specifically consider this case they don't write the non-negativity constraint into the Lagrangian, and i believe this effects the analysis at the end. I have attached a screenshot of the "answer."

UPDATE: The specific equation they derive in the screenshot actually goes on to be very important in the following question. So thoughts are very much appreciated, thank!

$\mathcal{L}(k, l, \lambda, \mu_k, \mu_l) = -rk - wl + \lambda [(k + 1)^\alpha l^{1 - \alpha} - y] + \mu_k k + \mu_l l$`

1. With respect to k:

$\frac{\partial \mathcal{L}}{\partial k} = -r + \lambda \alpha (k + 1)^{\alpha - 1} l^{1 - \alpha} + \mu_k$

2. With respect to l:

$\frac{\partial \mathcal{L}}{\partial l} = -w + \lambda (1 - \alpha) (k + 1)^\alpha l^{-\alpha} + \mu_l$

3. With respect to λ:

$\frac{\partial \mathcal{L}}{\partial \lambda} = (k + 1)^\alpha l^{1 - \alpha} - y$

Now, consider the specific case where $k = 0$ and $l > 0$:`

1. $\frac{\partial \mathcal{L}}{\partial k}\bigg|_{k=0} = -r + \lambda \alpha l^{1 - \alpha} + \mu_k \le 0$

2. $\frac{\partial \mathcal{L}}{\partial l}\bigg|_{k=0} = -w + \lambda (1 - \alpha) l^{-\alpha} = 0$

3. $\frac{\partial \mathcal{L}}{\partial \lambda}\bigg|_{k=0} = l^{1 - \alpha} - y = 0$

• $k^* = 0$
• $l^* = y^{\frac{1}{1-\alpha}}$
• $λ = \frac{w}{1-\alpha}y^{\frac{1}{1-\alpha}}$

1. They solve for for a condition on $y$ (see screenshot) consistent with $k = 0$ but their lack of an explicit use of a $\mu_k$ for our non-negativity on $k$ results in a different condition to what i would get, which would include $\mu_k$ . Why have they not explicitly used $\mu_k$ and $\mu_l$ as i have??
2. I also get a weird condition on $\mu_k$ by just substituting in $l^*$ and $λ^*$ does it looks right?

• $\mu_k \le r - \frac{\alpha wy^{\frac{2-\alpha}{1-\alpha}}}{1-\alpha}$

Screen Shot from Answer showing condition on k

Answer 1

You have made a mistake in your calculations. Your expression for $\lambda^*$ when $k=0$ is missing an $\alpha$. It should be $\lambda^*=\frac{w}{1-\alpha}y^{\frac{\alpha}{1-\alpha}}$ (as in the screenshot) rather than $\lambda^*=\frac{w}{1-\alpha}y^{\frac{1}{1-\alpha}}$.

Check your working from when you solved for $\lambda^*$ after substituting $l^*$ into equation 2 (when $k=0$).