<p>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.</p>

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2020-04-27T23:49:44.593Z
2020-04-27T23:49:44.593Z
VARulle
<p>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 <span class="math-container">$\geq 0$</span>, K <span class="math-container">$\geq 0$</span>.</p>

<p>The firm's constraint can be written as <span class="math-container">$wL + rK$</span> <span class="math-container">$\leq Q$</span></p>

<p>How does one prove that such a set is convex? We know that if a constraint is convex, some vector, z = <span class="math-container">$\lambda x + (1-\lambda)y$</span> lies within the set. </p>

<p>Suppose I choose 
<span class="math-container">$$z_1 = \lambda L_1 + (1-\lambda) L_2 $$</span></p>

<p><span class="math-container">$$z_2 = \lambda K_1 + (1-\lambda) K_2 $$</span></p>

<p>where <span class="math-container">$\lambda  \exists [0,1]$</span></p>

<p><strong>z</strong> = L + K</p>

<p><span class="math-container">$$wz_1 + rz_2 =w[\lambda L_1 + (1-\lambda) L_2] + r[\lambda K_1 + (1-\lambda) K_2] $$</span>
<span class="math-container">$$=\lambda[wL_1 + rK_1] +(1-\lambda)[wL_2 + rK_2]$$</span>
<span class="math-container">$$\leq \lambda Q + (1-\lambda)Q$$</span>
<span class="math-container">$$= Q$$</span></p>

<p>Thank you. </p>

<mathematical-economics>
How do I confirm that the constraint set wL + rK < Q is convex
CorporateNationalism
# How do I confirm that the constraint set wL + rK < Q is convex

(Asked by: username_0 on 2020-04-26)

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.

**Tags:** mathematical-economics

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## 1 answers

### Answer by username_1 on 2020-04-27. Score: 2

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.


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