# Selling inventory while maximizing the present value of profit

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 $$x_3$$.

• You did not even define $x_t$. – Rodrigo de Azevedo Apr 21 '19 at 7:33

## 1 Answer

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 $$x3 = 100 - x_1-x_2$$

• Could you please provide a detailed solution? – user 25 Apr 21 '19 at 7:23
• @RaashidShah I updated my answer with a detailed solution. Please don't forget to accept the answer if it helped! – EhsanK Apr 21 '19 at 15:28