1.2 Problem Statement: Given the total cost function $TC = 200 + 3Q$ and the demand function $P = 107 - 2Q$. 1. Write down the total revenue function. 2. Graph the total cost and total revenue functions on the same axes. 3. Estimate the break-even points both graphically and algebraically. --- 1. Total revenue (TR) is price times quantity: $$TR = P \times Q = (107 - 2Q)Q = 107Q - 2Q^2$$ 2. To find break-even points, set total revenue equal to total cost: $$TR = TC$$ $$107Q - 2Q^2 = 200 + 3Q$$ Rearrange: $$-2Q^2 + 107Q - 3Q - 200 = 0$$ $$-2Q^2 + 104Q - 200 = 0$$ Multiply both sides by $-1$ to simplify: $$2Q^2 - 104Q + 200 = 0$$ Divide entire equation by 2: $$Q^2 - 52Q + 100 = 0$$ 3. Solve quadratic equation using the quadratic formula: $$Q = \frac{52 \pm \sqrt{52^2 - 4 \times 1 \times 100}}{2}$$ Calculate discriminant: $$52^2 = 2704$$ $$4 \times 1 \times 100 = 400$$ $$\sqrt{2704 - 400} = \sqrt{2304} = 48$$ So, $$Q = \frac{52 \pm 48}{2}$$ Two solutions: $$Q_1 = \frac{52 + 48}{2} = \frac{100}{2} = 50$$ $$Q_2 = \frac{52 - 48}{2} = \frac{4}{2} = 2$$ 4. Break-even points are at $Q=2$ and $Q=50$. Calculate corresponding prices: At $Q=2$: $$P = 107 - 2(2) = 107 - 4 = 103$$ At $Q=50$: $$P = 107 - 2(50) = 107 - 100 = 7$$ 5. Graphically, the total cost line $TC = 200 + 3Q$ is a straight line starting at 200 when $Q=0$ with slope 3. The total revenue curve $TR = 107Q - 2Q^2$ is a downward-opening parabola. The break-even points are where these two graphs intersect at $Q=2$ and $Q=50$. --- Final answers: - Total revenue function: $TR = 107Q - 2Q^2$ - Break-even points: $Q=2$ and $Q=50$ with prices $P=103$ and $P=7$ respectively.