Suppose a firm produces Q units using labour (L) for which the wage, w>0 and the price of capital (K) is r>0. It cannot employ negative amounts of its factors: L $\geq 0$, K $\geq 0$.

The firm's constraint can be written as $wL + rK$ $\leq Q$

How does one prove that such a set is convex? We know that if a constraint is convex, some vector, z = $\lambda x + (1-\lambda)y$ lies within the set.

Suppose I choose $$z_1 = \lambda L_1 + (1-\lambda) L_2 $$

$$z_2 = \lambda K_1 + (1-\lambda) K_2 $$

where $\lambda \exists [0,1]$

z = L + K

$$wz_1 + rz_2 =w[\lambda L_1 + (1-\lambda) L_2] + r[\lambda K_1 + (1-\lambda) K_2] $$ $$=\lambda[wL_1 + rK_1] +(1-\lambda)[wL_2 + rK_2]$$ $$\leq \lambda Q + (1-\lambda)Q$$ $$= Q$$

Thank you.


Given the three inequalities you wrote down, the constraint set is the intersection of Quadrant I with a half-space in the 2-dimensional plane . Both these sets are convex. The intersection of convex sets is convex, so the constraint set is convex.


Consider an arbitrary pair of vectors $(L_{1},K_{1})$ and $(L_{2},K_{2})$ that satisfy $wL_{i} + rK_{i} \leq Q$ for $i = 1, 2$. To show that constraint set is convex, we need to show that any convex combination of that (arbitrary) pair of vectors lies in the constraint set. Namely, for all $\gamma \in [0,1]$ we have

$$\begin{split} w\left(\gamma L_{1} + (1-\gamma)L_{2}\right) + r\left(\gamma K_{1} + (1-\gamma)K_{2}\right) &= \gamma \left(w L_{1} + r K_{1}\right) + \left(1-\gamma \right)\left(w L_{2} + r K_{2}\right)\\ &\leq \gamma Q + (1-\gamma)Q = Q \end{split}$$ Thus the vector, $(L_{3},K_{3})$, is also in the constraint set (where $L_{3} := \gamma L_{1} + (1-\gamma)L_{2}$, and $K_{3} := \gamma K_{1} + (1-\gamma)K_{2}$) and so the set is convex.

Your answer (or at least the rearranging of inequalities in your answer) is largely correct. However, it is important to note that you've shown this inequality for an arbitrary vector in the set and for an arbitrary convex combination and not for a particular one.


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