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 artdition, 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. (5pts) Use your work in Problem 2 to find the marginal revenue $R(x)$. Then compute and interpret the following: $R^{\prime}(500000), R^{\prime}(1600000)$, and $R^{\prime}(2300000)$. Expression for $\mathrm{R}(\mathrm{x})=\mathrm{x}^{*}(6400-0.002 \mathrm{x})$
Solution
Step 1 ：Given the revenue function \(R(x) = x * (6400 - 0.002x)\), we need to find the derivative of this function to get the marginal revenue function.
Step 2 ：Using the power rule and the product rule for differentiation, the derivative of \(R(x)\) is \(R'(x) = 6400 - 0.004x\).
Step 3 ：Substitute \(x = 500000\) into \(R'(x)\) to get \(R'(500000) = 6400 - 0.004*500000 = 4400\).
Step 4 ：Substitute \(x = 1600000\) into \(R'(x)\) to get \(R'(1600000) = 6400 - 0.004*1600000 = 0\).
Step 5 ：Substitute \(x = 2300000\) into \(R'(x)\) to get \(R'(2300000) = 6400 - 0.004*2300000 = -2800\).
Step 6 ：Interpret the results: The marginal revenue at \(x = 500000\) is \$4400, at \(x = 1600000\) is \$0, and at \(x = 2300000\) is -\$2800. This means that at \(x = 500000\), each additional unit of output will increase the revenue by \$4400. At \(x = 1600000\), each additional unit of output will not change the revenue. At \(x = 2300000\), each additional unit of output will decrease the revenue by \$2800. This is because the marginal revenue is decreasing as the quantity of output increases, which is a common characteristic of diminishing returns.
Step 7 ：\(\boxed{R'(500000) = 4400, R'(1600000) = 0, R'(2300000) = -2800}\)
