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I am having a problem determining "which utility function is not like the rest." The following utility functions are:

$$U_1(x,y)=x^{2/3}y^{1/3}$$

$$U2(x,y)=3x^2*y+2$$

$$U_3(x,y)=\frac{1}{3}\ln (x)+\frac{2}{3}\ln( y)$$

$$U_4(x,y)= 4\ln(x)+2\ln(y)-12$$

So my general thought in approaching this question was to obtain the MRS for each of the functions. After doing this, however, I found U1 and U2 to have an MRS of -2y/x. However, I am having a much bigger problem differentiating U3 and U4. My intuition is that U4 would be equal to U1 and U2 because we value x twice as much as we value y. however, can anyone help out with the math?

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The rule for differentiating a natural logarithm is

$$\frac{d}{dx}a\ln(x)=\frac{a}{x}$$

so $$\begin{align}\frac{\partial{}}{\partial{}x}U_3(x,y)&=\frac{\partial{}}{\partial{}x}\left(\frac{1}{3}\ln(x)+\frac{2}{3}\ln(y)\right)\\&=\frac{1}{3}\frac{1}{x}\\&=\frac{1}{3x}.\end{align}$$

$$\begin{align}\frac{\partial{}}{\partial{}y}U_3(x,y)&=\frac{\partial{}}{\partial{}y}\left(\frac{1}{3}\ln(x)+\frac{2}{3}\ln(y)\right)\\&=\frac{2}{3}\frac{1}{y}\\&=\frac{2}{3y}.\end{align}$$

$$\text{MRS}=-\frac{\frac{\partial{}}{\partial{}x}U_3(x,y)}{\frac{\partial{}}{\partial{}y}U_3(x,y)}=-\frac{\frac{1}{3x}}{\frac{2}{3y}}=-\frac{y}{2x},$$

I guess that by following this template you should be able to work out the other derivatives and the MRSs that you are looking for.

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  • $\begingroup$ So when I take this partial derivative, my final answer here would look like this: (-1/3x+2lny)/(1/3lnx+2/3y)? I appreciate your help! $\endgroup$ – Zach Smith Feb 13 '16 at 17:00
  • $\begingroup$ @ZachSmith No, I finished off the example for $U_3$ so you can see how the whole process works. The method for calculating the MRS for $U_4$ is similar. $\endgroup$ – Ubiquitous Feb 13 '16 at 18:15

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