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Q: Suppose that all you do in a day is work, play and sleep. Let $x_1$ be the number of hours per day you spend playing, $x_2$ be the number of hours you spend sleeping, and $x_3$ be the number of hours you spend working. Suppose that sleeping is free, but playing costs you $17 an hour. Furthermore, you spend all the money you earn from working on playing.

The utility you get from sleeping and playing is given by a Cobb-Douglas utility function:

$U = x_1^{a_1} \times x_2^{a_2}$, where $a_1+a_2=1$.

If $a_1 = \frac 34$, and your hourly wage is $w$, find the number of hours you should work a day ($x_3$) in order to maximize your utility as a function of $w$.

I know Lagrange multiplier can be applied to this question.

So I write:

$L(x_1,x_2,x_3,λ_1,λ_2) = x_1^{a_1} \times x_2^{a_2} + λ_1(17x_1 - w x_3) - λ_2(x_1 + x_2 + x_3 -24),$

where the first function is $U$, the second one is the assumption that you spend all the money you earn from working on playing.

Solving gives: $6x_1 = x_2, x_1 = w \frac{x_3}{17}$, when $U$ is maximized.

Thus, $x_3 = \frac{408}{7w+17}$.

However, this is wrong.

So, is my Lagrange function incorrect?

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  • $\begingroup$ You've made a typo in the Lag-mult function? $\lambda_1(17x_1-wx_3)\dots$ $\endgroup$ – Parcly Taxel Oct 15 '18 at 11:56
  • $\begingroup$ Thank you. I've edited it. $\endgroup$ – lamhei Oct 15 '18 at 12:18
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$$ L(x,\lambda) = x_1^{\frac 34}x_2^{\frac 14}+\lambda_1(w x_3-17 x_1)+\lambda_2(24-x_1-x_2-x_3) $$

$$ \nabla L = \left\{\begin{array}{rcl} \frac{3 \sqrt[4]{\text{x2}}}{4 \sqrt[4]{\text{x1}}}-17 \lambda_1-\lambda_2&=&0 \\ \frac{\text{x1}^{3/4}}{4 \text{x2}^{3/4}}-\lambda_2&=&0 \\ \lambda_1 w-\lambda_2&=&0 \\ w x_3-17 x_1&=&0 \\ 24-x_1-x_2-x_3&=&0 \\ \end{array}\right. $$

with the solution

$$ \left[ \begin{array}{cccccc} x_1 & x_2 & x_3 & \lambda_1 & \lambda_2 & U\\ \frac{18 w}{w+17} & 6 & \frac{306}{w+17} & -\frac{3^{3/4}}{4 \sqrt[4]{w} \sqrt[4]{(w+17)^3}} & -\frac{3^{3/4} w^{3/4}}{4 \sqrt[4]{(w+17)^3}} & 6\ 3^{3/4} \left(\frac{w}{w+17}\right)^{3/4} \end{array} \right] $$

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