Let's solve the problem step-by-step: ### Given: - **Demand Function:** \( P = 20 - 2Q \) - **Cost Function:** \( C = Q^2 + 6 \) --- ## (a) **Give the revenue and profit functions.** **Revenue \( R \):** \[ R = P \times Q \] Substitute \( P = 20 - 2Q \): \[ R = (20 - 2Q)Q = 20Q - 2Q^2 \] **Profit \( \pi \):** \[ \pi = R - C = (20Q - 2Q^2) - (Q^2 + 6) \] \[ \pi = 20Q - 2Q^2 - Q^2 - 6 \] \[ \boxed{\pi = 20Q - 3Q^2 - 6} \] --- ## (b) **Calculate the value of \( Q \) for which the business breaks even.** Break-even: Profit = 0 \[ 20Q - 3Q^2 - 6 = 0 \] \[ 3Q^2 - 20Q + 6 = 0 \] Use the quadratic formula: \[ Q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 3 \), \( b = -20 \), \( c = 6 \): \[ Q = \frac{20 \pm \sqrt{(-20)^2 - 4 \times 3 \times 6}}{2 \times 3} \] \[ Q = \frac{20 \pm \sqrt{400 - 72}}{6} \] \[ Q = \frac{20 \pm \sqrt{328}}{6} \] \[ \sqrt{328} \approx 18.11 \] \[ Q = \frac{20 \pm 18.11}{6} \] So, \[ Q_1 = \frac{20 + 18.11}{6} \approx \frac{38.11}{6} \approx 6.35 \] \[ Q_2 = \frac{20 - 18.11}{6} \approx \frac{1.89}{6} \approx 0.315 \] **Break-even quantities:** \[ \boxed{Q \approx 0.32 \text{ and } Q \approx 6.35} \] --- ## (c) **Calculate the value of \( Q \) that would result in the maximal profit possible, and the value of this profit.** Maximize \( \pi(Q) = 20Q - 3Q^2 - 6 \). Take the derivative and set to zero: \[ \frac{d\pi}{dQ} = 20 - 6Q = 0 \] \[ 6Q = 20 \implies Q = \frac{20}{6} = 3.33 \] **Maximum profit:** \[ \pi(3.33) = 20(3.33) - 3(3.33)^2 - 6 \] \[ = 66.67 - 3 \times 11.11 - 6 \] \[ = 66.67 - 33.33 - 6 \] \[ = 66.67 - 39.33 \] \[ = 27.34 \] --- ## **Final Answers (summary):** ### (a) Revenue and profit functions: - \( R = 20Q - 2Q^2 \) - \( \pi = 20Q - 3Q^2 - 6 \) ### (b) Break-even values of \( Q \): - \( Q \approx 0.32 \) and \( Q \approx 6.35 \) ### (c) Maximum profit and corresponding \( Q \): - \( Q = 3.33 \) - Maximum profit \( \approx 27.34 \)