Maximum of a utility function

Question

I was practising some related rates problems and I came across this particularly difficult question. The question goes like this:

Suppose Peyam's utility function is given by:

$U(\theta)=\frac{\mu C}{\mu\sin(\theta)\;+\;\cos(\theta)}$

where C is a universal cake-constant and $\mu$ is the intensity of happiness.

Note: Think of $\theta$ here as the time of the year. So $\theta=0$ corresponds to the beginning of January 1st, and $\theta=2\pi$ corresponds to the end of December 31st.

Show that Peyam is happiest precisely when $\tan(\theta)=\mu$.

I believe what is being asked for is the value of the maximum value of $\mu$, which should be equal to $\tan(\theta)$

I tried by making $\mu$ the subject, and ended up with

$\mu=\frac{U(\theta)\cos(\theta)}{C-U(\theta)\sin(\theta)}$

I'm guessing the next thing to do is to differentiate, but I'm not exactly sure how to differentiate $U(\theta)$. Do I substitute back in the original equation for $U(\theta)$? How can I move on from here?

Answer 1

It looks like we want to treat the "utility" as a number $ \ 0 \ < \ U \ < \ 1 \ $ in order to keep the ratio defined "everywhere". [Graph the curve $ \ \frac{y}{y·\sin(x) \ + \ \cos(x)} $ equal to a constant and vary the constant's value to see this.] There is then a "family" of such utility functions which have maxima in the specified interval for $ \ \theta \ . $

Since $ \ \mu \ $ is a variable in addition to $ \ \theta \ , \ $ it will help to write the relation as $$ U·(\mu \sin \theta \ + \ \cos \theta) \ \ = \ \ \mu · C \ \ , $$

which we would differentiate implicitly with respect to $ \ \theta \ \ , $ thus $$ \frac{d}{d \theta} \ [ \ U · (\mu · \sin \theta \ + \ \cos \theta) \ ] \ \ = \ \ \frac{d}{d \theta} \ [ \ \mu · C \ ] $$ $$ \Rightarrow \ \ U · \left( \frac{d \mu}{d \theta} · \sin \theta \ + \mu·\cos \theta \ - \ \sin \theta \right) \ \ = \ \ C · \frac{d \mu}{d \theta} $$ $$ \Rightarrow \ \ \frac{d \mu}{d \theta} · (C \ - \ U · \sin \theta) \ \ = \ \ U · ( \mu·\cos \theta \ - \ \sin \theta ) \ \ . $$

Since we want the maximal value for $ \ \mu \ $ in any particular utility function, we would set $ \ \frac{d \mu}{d \theta} \ = \ 0 \ \ $ and the result for $ \ \mu \ $ follows. So this is sort of a "related-rates" problem...

Answer 2

No need to make $\mu$ a subject.

You take derivative of the utility function and equate to zero: $$U'(\theta)=\left[\frac{\mu C}{\mu\sin(\theta)\;+\;\cos(\theta)}\right]'=-\frac{\mu C\cdot [\mu \cos(\theta)-\sin(\theta)]}{\mu \sin(\theta)+\cos(\theta)}=0 $$ Can you finish it yourself?