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A company owns an inventory of $100$ units of a good. It must sell the entire inventory over the next three periods. The profit function for sales within any one period is

$$\pi(x_t) = 50x_t − 0.5x^2_t, \qquad t \in \{1, 2, 3\}$$

It wishes to maximize its present value of profit

$$V = \pi(x_1) + \beta \pi(x_2) + \beta^2 \pi(x_3)$$

where $\beta = 0.8$ is the discount factor. Find the optimal values of $x_1$, $x_2$, and $x_3$.

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  • $\begingroup$ You did not even define $x_t$. $\endgroup$ – Rodrigo de Azevedo Apr 21 at 7:33
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Your $π(x_t) = 50x_t − 0.5x_t^2$ is a concave function, thus, your profit $V$ function is also concave. The variables in your problem are $x$. So, you should be able to get the maximal profit using derivative over your $x$s.

Detailed:

  1. $V = 50x_1 − 0.5x_1^2 + 0.8(50x_2 − 0.5x_2^2) + 0.64(50x_3 − 0.5x_3^2)$
  2. $x_1 +x_2 +x_3 =100 \Rightarrow x_3 = 100 - x_1 - x_2$
  3. substitute $x_3$ in the above equation for $V$. You will get:

$$V=50x_1 - 0.5x_1^2 + 0.8(50x_2 - 0.5x_2^2)+ 0.64(50(100-x_1-x_2)- 0.5(100-x_1-x_2)^2) $$

  1. $V$ is concave so you can get to the maximum point taking partial derivative of the variables and equaling them to zero:

$$\frac{\partial V} {\partial x_1} = 50 - x_1 -32 +0.64(100-x_1-x_2)=0\\ \frac{\partial V} {\partial x_2} = 40 - 0.8x_2 -32 +0.64(100-x_1-x_2)=0 $$ Now you just have a $2\times 2$ system of equation. Solve it (like this) and you get your optimal value for $x_1, x_2$. and you get $x3 = 100 - x_1-x_2$

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  • $\begingroup$ Could you please provide a detailed solution? $\endgroup$ – user 25 Apr 21 at 7:23
  • $\begingroup$ @RaashidShah I updated my answer with a detailed solution. Please don't forget to accept the answer if it helped! $\endgroup$ – EhsanK Apr 21 at 15:28

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