Coffee Profit 0F830A

1. **State the problem:** We have two coffee blends, Blend A and Blend B, with demand functions and a cost function. We want to find the total revenue, total profit functions, and determine the maximum profit along with the quantities and prices of both blends. 2. **Total Revenue (TR):** Given demand functions: $$P_A(x) = 45 - 0.05x$$ $$P_B(y) = 60 - 0.1y$$ Total revenue is: $$TR = P_A(x) \times x + P_B(y) \times y = (45 - 0.05x)x + (60 - 0.1y)y$$ Simplify: $$TR = 45x - 0.05x^2 + 60y - 0.1y^2$$ 3. **Total Cost (C):** Given cost function: $$C(x,y) = 15x + 10y + 0.02xy + 800$$ 4. **Profit function (\(\pi\))**: Profit is revenue minus cost: $$\pi = TR - C = (45x - 0.05x^2 + 60y - 0.1y^2) - (15x + 10y + 0.02xy + 800)$$ Simplify: $$\pi = 30x - 0.05x^2 + 50y - 0.1y^2 - 0.02xy - 800$$ 5. **Find critical points by setting partial derivatives to zero:** Partial derivatives: $$\pi_x = \frac{\partial \pi}{\partial x} = 30 - 0.1x - 0.02y$$ $$\pi_y = \frac{\partial \pi}{\partial y} = 50 - 0.2y - 0.02x$$ Set to zero: $$0 = 30 - 0.1x - 0.02y \implies 0.1x = 30 - 0.02y \implies x = \frac{30 - 0.02y}{0.1} = 300 - 0.2y$$ Substitute into \(\pi_y=0\): $$0 = 50 - 0.2y - 0.02(300 - 0.2y)$$ Simplify: $$0 = 50 - 0.2y - 6 + 0.004y = 44 - 0.196y$$ $$0.196y = 44 \implies y = \frac{44}{0.196} \approx 224.49$$ Recalculate with original values (from user): $$0 = 50 - 0.2y - 0.02(300 - 0.2y)$$ $$0 = 50 - 0.2y - 6 + 0.004y = 44 - 0.196y$$ $$y = \frac{44}{0.196} = 224.49$$ Then, $$x = 300 - 0.2(224.49) = 300 - 44.898 = 255.102$$ 6. **Second derivative test:** $$\pi_{xx} = -0.1 < 0$$ $$\pi_{yy} = -0.2 < 0$$ $$\pi_{xy} = -0.02$$ Discriminant: $$D = \pi_{xx} \pi_{yy} - (\pi_{xy})^2 = (-0.1)(-0.2) - (-0.02)^2 = 0.02 - 0.0004 = 0.0196 > 0$$ Since \(D > 0\) and \(\pi_{xx} < 0\), the critical point is a maximum. 7. **Calculate maximum profit:** $$\pi = 30x - 0.05x^2 + 50y - 0.1y^2 - 0.02xy - 800$$ Substitute \(x=255.102, y=224.49\): $$\pi = 30(255.102) - 0.05(255.102)^2 + 50(224.49) - 0.1(224.49)^2 - 0.02(255.102)(224.49) - 800$$ Calculate each term: $$30 \times 255.102 = 7653.06$$ $$0.05 \times (255.102)^2 = 0.05 \times 65085.99 = 3254.3$$ $$50 \times 224.49 = 11224.5$$ $$0.1 \times (224.49)^2 = 0.1 \times 50398.4 = 5039.84$$ $$0.02 \times 255.102 \times 224.49 = 0.02 \times 57256.5 = 1145.13$$ Sum: $$\pi = 7653.06 - 3254.3 + 11224.5 - 5039.84 - 1145.13 - 800 = 7638.29$$ 8. **Find prices at maximum profit:** $$P_A = 45 - 0.05x = 45 - 0.05(255.102) = 45 - 12.755 = 32.245$$ $$P_B = 60 - 0.1y = 60 - 0.1(224.49) = 60 - 22.449 = 37.551$$ **Final answers:** - Maximum profit \(\approx 7638.29\) - Blend A quantity \(x \approx 255.1\) kg, price \(P_A \approx 32.25\) - Blend B quantity \(y \approx 224.5\) kg, price \(P_B \approx 37.55\)