<p>The utility function is $U(C,R)=CR$ and the time is restricted: $48=R+L$. Now we know that $C=\sqrt L$. $C$ can be replaced by $\sqrt L$. Therefore the langrarian is</p>

<p>$\mathcal L=\sqrt L\cdot R+\lambda (48-L-R)$</p>

<p>The (partial) derivatives are the following. They have to be set equal to zero.</p>

<p>$\frac{\partial \mathcal L}{\partial L}=\frac12 L^{-0.5} R-\lambda=0$</p>

<p>$\frac{\partial \mathcal L}{\partial R}= L^{0.5} -\lambda=0$</p>

<p>Putting $\lambda$ on the RHS</p>

<p>$\frac12 L^{-0.5} R=\lambda \quad (1)$</p>

<p>$L^{0.5} =\lambda \quad (2)$</p>

<p>Dividing (1) by (2):</p>

<p>$\frac12\cdot \frac{R}{L} =1 \Rightarrow R=2L$</p>

<p>The expression for R can be insert in the time restriction</p>

<p>$48=2L+L$</p>

CC BY-SA 3.0
2016-03-29T10:14:21.800Z
That´s fine. You´re welcome.
callculus42
CC BY-SA 3.0
2016-03-28T18:38:39.740Z
2016-03-28T18:38:39.740Z
callculus42
<p>Robinson Crusoe’s preferences over coconut consumption, C, and leisure, R, are represented by the utility function U(C, R) = CR. There are 48 hours available for Robinson to allocate between labor and leisure. If he works L hours, he will produce the square root of L of coconuts. He will choose to work. </p>

<p>The answer is 16 and I known this but confused to its working out. Can some give me a step by step?</p>

<microeconomics><mathematical-economics><production-function>
Robinson Crusoe Production Economy
Linsky
# Robinson Crusoe Production Economy

(Asked by: username_0 on 2016-03-28)

Robinson Crusoe’s preferences over coconut consumption, C, and leisure, R, are represented by the utility function U(C, R) = CR. There are 48 hours available for Robinson to allocate between labor and leisure. If he works L hours, he will produce the square root of L of coconuts. He will choose to work.

The answer is 16 and I known this but confused to its working out. Can some give me a step by step?

**Tags:** microeconomics, mathematical-economics, production-function

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## 1 answers

### Answer by username_1 on 2016-03-28. Score: 1 | Accepted answer

The utility function is $U(C,R)=CR$ and the time is restricted: $48=R+L$. Now we know that $C=\sqrt L$. $C$ can be replaced by $\sqrt L$. Therefore the langrarian is

$\mathcal L=\sqrt L\cdot R+\lambda (48-L-R)$

The (partial) derivatives are the following. They have to be set equal to zero.

$\frac{\partial \mathcal L}{\partial L}=\frac12 L^{-0.5} R-\lambda=0$

$\frac{\partial \mathcal L}{\partial R}= L^{0.5} -\lambda=0$

Putting $\lambda$ on the RHS

$\frac12 L^{-0.5} R=\lambda \quad (1)$

$L^{0.5} =\lambda \quad (2)$

Dividing (1) by (2):

$\frac12\cdot \frac{R}{L} =1 \Rightarrow R=2L$

The expression for R can be insert in the time restriction

$48=2L+L$

> **Comments:**
> **username_1**: That´s fine. You´re welcome.  

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