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) 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 $R(x)$.

Answer

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Answer

\(\boxed{\text{Final Answer: The expression for the price } p \text{ in terms of the demand } x \text{ is } p = 6400 - 0.002x. \text{ The domain of the price-demand equation is } x \geq 0 \text{ and } x \leq 3200000.}\)

Steps

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 :\(\boxed{\text{Final Answer: The expression for the price } p \text{ in terms of the demand } x \text{ is } p = 6400 - 0.002x. \text{ The domain of the price-demand equation is } x \geq 0 \text{ and } x \leq 3200000.}\)

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