Problem
Introduction: Production, Price, Demand, Revenue, \& Profit A technology startup's market research department is tasked with determining the market viability of a new smartphone device. After suitable testing on the interest in a new smartphone, the research department determines the following price-demand equation: \[ x=3.2 \times 10^{6}-500 p \] where $x$ is the amount of units (smartphones) in demand at price $p$ (in dollars). For example, if the price of the new smartphone is set at $p=\$ 100$, then the amount of new smartphones in demand should be: \[ x=3.2 \times 10^{6}-500(100)=3150000 \text { units } \] In addition, the financial department provides the cost function measured in dollars: \[ C(x)=85 x+50000 \] where $x$ is the number of smartphones produced. Note that $\$ 50000$ is the fixed costs of production (maintenance, overhead, etc.) and $\$ 85$ is the cost (labor, materials, marketing, transportation, storage, etc.) per smartphone. 1. (10pts) Assume that the startup will be able to sell all smartphones produced. The revenue function $R(x)$ can be described in words as: \[ R(x)=\text { (number of smartphones sold)(the price per smartphone) } \] Use your work in Problem 1 to find the expression for $R(x)$.
Solution
Step 1 ：The revenue function is the product of the number of units sold and the price per unit. In this case, the number of units sold is represented by \(x\) and the price per unit is given by the equation \(p = 3.2 \times 10^{6} - 500x\). Therefore, the revenue function \(R(x)\) can be expressed as \(R(x) = x \cdot (3.2 \times 10^{6} - 500x)\).
Step 2 ：Substitute \(x\) and \(p\) into the equation to get \(R = x*(3200000.0 - 500*x)\).
Step 3 ：\(\boxed{R(x) = x \cdot (3.2 \times 10^{6} - 500x)}\) is the final answer.
