Expected profit maximization

Question

Each concert a singer makes profit of 0.75, but after every concert a singert can fall into bad mood with probability = 0.5. To get a singer out of these mood producer can send her flowers. If flowers cost $x$ money, then singer can get out of bad mood with probability $\sqrt{x}$. Which $x$ should producer choose in order to maximize his expected profit. (All the profit from concert goes to producer. One can make assumprion, that singer's career lasts $n$ days.)

Answer 1

I'm not sure that this is the correct path, but one can solve it by brute force using backward induction.

Let $f(n)$ be the expected profit when there are $n$ days left and the singer is happy today and $g(n)$ be the expected profit when there are $n$ days left and the singer is sad today. I assume that if she's sad, she stays sad until flowers are sent and make her happy and there is one concert a day.

Clearly, $f(1)=0.75$ and $g(1)=\max\limits_{0\leq x\leq 1} \left\{\sqrt{x}0.75-x\right\}=\tfrac{9}{64}$.

For every $n>1$, $f(n)=0.75+\tfrac{f(n-1)+g(n-1)}{2}$ (no need for flowers, she's happy!) and $g(n)=\max\limits_{0\leq x\leq 1} \left\{\sqrt{x}\left[0.75 + \tfrac{f(n-1)+g(n-1)}{2}\right]+(1-\sqrt{x})g(n-1)-x\right\}$.

I tried to calculate it for $n=2,3$ but it got ugly pretty quickly so I'm not sure there is a closed-form solution that can be guessed using this way.

Hope it helps!