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) Find the expression for the price $p$ in terms of the demand $x$ from the price-demand equation shown in the introduction. Then, determine the domain of the price-demand equation from your work. Note that both price $p$ and demand $x$ must be non-negative (greater than or equal to zero). 2. (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 Problem1 to find the expression for $R(x)$.
Solution
Step 1 ：Rearrange the equation \(x=3.2 \times 10^{6}-500 p\) to solve for \(p\).
Step 2 ：Subtract \(x\) from both sides of the equation to get \(-500p = -x + 3.2 \times 10^{6}\).
Step 3 ：Divide both sides of the equation by \(-500\) to isolate \(p\) on one side of the equation. This gives us \(p = 6400 - 0.002x\).
Step 4 ：Determine the domain of the price-demand equation. Since both price \(p\) and demand \(x\) must be non-negative, this means \(x\) must be greater than or equal to zero.
Step 5 ：Also, since the maximum value of \(x\) is when \(p=0\), we substitute \(p=0\) into the equation to get \(x = 3.2 \times 10^{6}\).
Step 6 ：So, the domain of the price-demand equation is \(x \geq 0\) and \(x \leq 3200000\).
Step 7 ：\(\boxed{\text{Final Answer: The expression for the price } p \text{ in terms of the demand } x \text{ is } p = 6400 - 0.002x. \text{ The domain of the price-demand equation is } x \geq 0 \text{ and } x \leq 3200000.}\)
