# Constrained maximization problem with linear function

I have the following problem. Given this function

$$E[\pi] = (1-r)[\alpha b- (1-p)C-K]+T$$

I would like to find the maximum w.r.t. $$r$$ given this constraint:

$$U = (1-r)b-T \geq 0$$.

It is an economic problem that I am formalizing, but this is not relevant to its solution, I am only interested in the mathematical resolution of the problem. I provide some background. Our variable $$r$$ is a number between $$0$$ and $$1$$, $$p$$ is some probability, $$\alpha,b,C,K$$ and $$T$$ are all positive constants. If it is useful for the resolution of the problem, we can also assume that $$\alpha$$ is between $$0$$ and $$1$$. An important assumption (namely, assumption $$\&$$) is that $$(1-p)C+K>\alpha b$$. Readers used to economics may recognize that the objective function is a sort of expected profit and the constraint is a sort of expected utility. I tried with Kuhn-Tucker, but with miserable results since deriving w.r.t. $$r$$ does not yield any expression with $$r$$.

Now I'm following a more intuitive approach. I start by assuming that $$U=0$$ is the constraint, then I can find an expression for $$r$$ from the constraint and I substitute it in the target function. After easy steps, I get this

$$T \frac{\alpha b - (1-p)C-K}{b} + T$$.

At this point, I can use assumption $$\&$$ to conclude that the first $$T$$ above will be negative, but I'm not able to conclude whether it will be lower, equal or higher to/than the second $$T$$ because it is multiplied by some constant.

At this point I'm stuck. I do not know if my approach can work. Could you please suggest a nicer way to proceed? I am on the right track or it's a dead end?

For the admin: This is a new post since the old one was blocked because it lacked many information. I hope that in its current form it is acceptable to start a discussion. Let me know if I need to add anything else. Thank you.

Hint.

Calling $$\phi = (1-p)C+K-\alpha b\gt 0$$ and considering

$$U = (1-r)b-T \geq 0\Rightarrow r \le 1-\frac Tb$$

we can formulate the problem as

$$\min_r (r-1)\phi+T,\ \ \text{s. t.}\ \ \cases{r\ge 0\\ r\le1-\frac Tb}$$

now as $$(r-1)\phi+T$$ is linear, the solution is one of the set

$$\left\{T-\phi,T-\phi\frac Tb\right\}$$

• Using your hint here's my attempt (I stack with the max problem). $\max_r -(1-r) \phi+T$, in order to maximize, I have to minimize $(1-r)$. Given the constraint, $r=1-T/b$ at least. If $T/b \leq 1$, then $r=1-T/b$ is the solution. If $T/b>1$, then the problem has no solution because I cannot respect the constraint given that $r \in [0,1]$. Am I right? Say that $T/b \leq 1$, can I say something on the sign of $E[\pi]$ evaluated on $r=1-T/b$ with these data? Jun 10 '21 at 15:02