Question: 5) A company has started selling a new type of smartphone at the price of $120 - 0.05x$ where \( x…

5. A company has started selling a new type of smartphone at the price of $120 - 0.05x$ where \( x \) is the number of smartphones manufactured per day. The parts for each smartphone cost $60 and the labor and overhead for running the plant cost $6000 per day. How many smartphones should the company manufacture and sell per day to maximize profit?

Solution

To solve the problem, we need to compute the profit function and find its maximum. The price function is given by \[ P(x) = 120 - 0.05x \] The revenue function is the price per smartphone times the number of smartphones: \[ R(x) = x \cdot P(x) = x(120 - 0.05x) = 120x - 0.05x^2 \] The cost function is the sum of the cost of parts and the fixed cost: \[ C(x) = 60x + 6000 \] The profit function is the revenue minus the cost: \[ \pi(x) = R(x) - C(x) = (120x - 0.05x^2) - (60x + 6000) \] Simplify the profit function: \[ \pi(x) = 120x - 0.05x^2 - 60x - 6000 \] Combine like terms: \[ \pi(x) = -0.05x^2 + 60x - 6000 \] To find the number of smartphones to maximize the profit, take the derivative of \(\pi(x)\) with respect to \(x\) and set it to 0: \[ \pi’(x) = -0.1x + 60 = 0 \] Solve for \(x\): \[ -0.1x + 60 = 0 \] \[ -0.1x = -60 \] \[ x = 600 \] The company should manufacture and sell 600 smartphones per day to maximize profit.