1. **State the problem:** We have two coffee blends, Blend A and Blend B, with demand functions for price and a cost function. We want to find the total revenue, total profit functions, and then determine the maximum profit along with the quantities and prices of both blends. 2. **Write down the given functions:** - Price of Blend A: $P_A(x) = 45 - 0.05x$ - Price of Blend B: $P_B(y) = 60 - 0.1y$ - Cost function: $C(x,y) = 15x + 10y + 0.02xy + 800$ 3. **Total Revenue (TR):** Total revenue is price times quantity for each blend, summed: $$ TR = xP_A(x) + yP_B(y) = x(45 - 0.05x) + y(60 - 0.1y) $$ Simplify: $$ TR = 45x - 0.05x^2 + 60y - 0.1y^2 $$ 4. **Total Profit ($\Pi$) function:** Profit is total revenue minus total cost: $$ \Pi(x,y) = TR - C = (45x - 0.05x^2 + 60y - 0.1y^2) - (15x + 10y + 0.02xy + 800) $$ Simplify: $$ \Pi(x,y) = 45x - 0.05x^2 + 60y - 0.1y^2 - 15x - 10y - 0.02xy - 800 $$ $$ \Pi(x,y) = (45x - 15x) + (60y - 10y) - 0.05x^2 - 0.1y^2 - 0.02xy - 800 $$ $$ \Pi(x,y) = 30x + 50y - 0.05x^2 - 0.1y^2 - 0.02xy - 800 $$ 5. **Find critical points by setting partial derivatives to zero:** Calculate partial derivatives: $$ \frac{\partial \Pi}{\partial x} = 30 - 0.1x - 0.02y = 0 $$ $$ \frac{\partial \Pi}{\partial y} = 50 - 0.2y - 0.02x = 0 $$ 6. **Solve the system:** From first equation: $$ 30 - 0.1x - 0.02y = 0 \implies 0.1x + 0.02y = 30 $$ From second equation: $$ 50 - 0.2y - 0.02x = 0 \implies 0.02x + 0.2y = 50 $$ Rewrite system: $$ 0.1x + 0.02y = 30 $$ $$ 0.02x + 0.2y = 50 $$ Multiply second equation by 5 to align coefficients: $$ 0.1x + 0.02y = 30 $$ $$ 0.1x + y = 250 $$ Subtract first from second: $$ (0.1x + y) - (0.1x + 0.02y) = 250 - 30 $$ $$ y - 0.02y = 220 \implies 0.98y = 220 \implies y = \frac{220}{0.98} \approx 224.49 $$ Substitute $y$ back into first equation: $$ 0.1x + 0.02(224.49) = 30 \implies 0.1x + 4.4898 = 30 \implies 0.1x = 25.5102 \implies x = 255.10 $$ 7. **Calculate prices at these quantities:** $$ P_A = 45 - 0.05(255.10) = 45 - 12.755 = 32.245 $$ $$ P_B = 60 - 0.1(224.49) = 60 - 22.449 = 37.551 $$ 8. **Calculate maximum profit:** $$ \Pi(255.10, 224.49) = 30(255.10) + 50(224.49) - 0.05(255.10)^2 - 0.1(224.49)^2 - 0.02(255.10)(224.49) - 800 $$ Calculate each term: $$ 30 \times 255.10 = 7653 $$ $$ 50 \times 224.49 = 11224.5 $$ $$ 0.05 \times (255.10)^2 = 0.05 \times 65087.01 = 3254.35 $$ $$ 0.1 \times (224.49)^2 = 0.1 \times 50400.6 = 5040.06 $$ $$ 0.02 \times 255.10 \times 224.49 = 0.02 \times 57256.6 = 1145.13 $$ Sum positive terms: $$ 7653 + 11224.5 = 18877.5 $$ Sum negative terms: $$ 3254.35 + 5040.06 + 1145.13 + 800 = 10239.54 $$ Profit: $$ 18877.5 - 10239.54 = 8637.96 $$ **Final answers:** - Total revenue function: $TR = 45x - 0.05x^2 + 60y - 0.1y^2$ - Total profit function: $\Pi = 30x + 50y - 0.05x^2 - 0.1y^2 - 0.02xy - 800$ - Maximum profit approximately $8637.96$ - Optimal quantities: $x \approx 255.10$ kg of Blend A, $y \approx 224.49$ kg of Blend B - Prices at maximum profit: $P_A \approx 32.25$, $P_B \approx 37.55$