Let's go through the solution step by step: ### Given: The profit \( P \) (in dollars) from selling \( x \) bags of popcorn is: \[ P = 2.38x - \frac{x^2}{20000} - 3900 \] for \( 0 \leq x \leq 50000 \). --- ## (a) Intervals of Increase and Decrease ### Step 1: Find the derivative of \( P \) with respect to \( x \) \[ \frac{dP}{dx} = 2.38 - \frac{2x}{20000} \] \[ \frac{dP}{dx} = 2.38 - \frac{x}{10000} \] --- ### Step 2: Set the derivative equal to zero to find critical points \[ 2.38 - \frac{x}{10000} = 0 \] \[ \frac{x}{10000} = 2.38 \] \[ x = 2.38 \times 10000 = 23800 \] --- ### Step 3: Test intervals around the critical point - For \( x < 23800 \): \[ \frac{dP}{dx} = 2.38 - \frac{x}{10000} > 0 \] So, \( P \) is **increasing**. - For \( x > 23800 \): \[ \frac{dP}{dx} = 2.38 - \frac{x}{10000} < 0 \] So, \( P \) is **decreasing**. --- ### Step 4: Write intervals in notation - **Increasing on:** \( [0, 23800) \) - **Decreasing on:** \( (23800, 50000] \) --- ## (b) Maximum Profit The maximum profit occurs at the critical point where the derivative changes from positive to negative: \( x = 23800 \). **You should charge the price that yields sales of \( x = 23800 \) bags of popcorn.** This is because the profit function changes from increasing to decreasing at this value, so the profit is maximized when \( x = 23800 \). --- ## **Final Answers** ### (a) - **Increasing:** \( [0, 23800) \) - **Decreasing:** \( (23800, 50000] \) ### (b) - **Maximum profit at:** \( x = 23800 \) bags of popcorn. - **Reason:** The profit function changes from increasing to decreasing at this value, so maximum profit occurs here. --- If you need the actual maximum profit value, substitute \( x = 23800 \) back into the original \( P \) function: \[ P = 2.38(23800) - \frac{(23800)^2}{20000} - 3900 \]