Q. 5.6 - A First Course in Probability

Question

Your company must make a sealed bid for a construction project. If you succeed in winning the contract (by having the lowest bid), then you plan to pay another firm $\$100,000$ to do the work. If you believe that the minimum bid (in thousands of dollars) of the other participating companies can be modeled as the value of a random variable that is uniformly distributed on $(70, 140)$, how much should you bid to maximize your expected profit? 

Step-by-Step Solution

Answer

The maximum profit to be expected is $40/7$ thousands of dollars

Step1: Evaluate E ( P )

The lowest bid's density function is

$f\left(x\right)=\frac{1}{140-70}$

Otherwise, it is equal to zero for $x\in (70,140)$. Let's say we invest $x$ thousand dollars in our offer. We will make a profit of $(x-10)$ thousand dollars if our bid wins the competition. In that instance, the profit that can be expected is

$$\begin{gathered} E\left(P\right)=(x-100)\times \frac{140-x}{70} \\ =\frac{1}{70}\left(240x-x^2-14000\right) \end{gathered}$$

Step2: Find Maximum profit

Let's discover a value for $x$ that optimizes profit. We have that with distinguishing.

$$\begin{gathered} \frac{d}{dx}E\left(P\right)=\frac{1}{70}(240-2x) \\ =0 \end{gathered}$$

This means that $x=120$. As a result, the highest profit

$$\begin{gathered} {\left.E\left(P\right)\right|}_{x=120}=(120-100)\times \frac{140-120}{70} \\ =\frac{40}{7} \end{gathered}$$

thousand of dollars.