1. Problem: Calculate total raw materials required and total cost for products A and B. Given matrix of materials per unit: $$M=\begin{bmatrix}2 & 3 \\ 4 & 2 \\ 1 & 5\end{bmatrix}$$ Quantity vector: $$Q=\begin{bmatrix}150 \\ 200\end{bmatrix}$$ Step 1: Total materials required is matrix multiplication: $$M \times Q = \begin{bmatrix}2 & 3 \\ 4 & 2 \\ 1 & 5\end{bmatrix} \times \begin{bmatrix}150 \\ 200\end{bmatrix} = \begin{bmatrix}2\times150 + 3\times200 \\ 4\times150 + 2\times200 \\ 1\times150 + 5\times200\end{bmatrix} = \begin{bmatrix}300 + 600 \\ 600 + 400 \\ 150 + 1000\end{bmatrix} = \begin{bmatrix}900 \\ 1000 \\ 1150\end{bmatrix}$$ Step 2: Cost per kg vector: $$C = \begin{bmatrix}5 \\ 4 \\ 3\end{bmatrix}$$ Step 3: Total cost is dot product of total materials and cost per kg: $$\text{Total Cost} = 900\times5 + 1000\times4 + 1150\times3 = 4500 + 4000 + 3450 = 11950$$ --- 2. Problem: Compute net shipments after returns and revenue. Given shipments matrix: $$A=\begin{bmatrix}120 & 80 & 60 \\ 90 & 50 & 70\end{bmatrix}$$ Returns matrix: $$R=\begin{bmatrix}5 & 2 & 1 \\ 3 & 4 & 2\end{bmatrix}$$ Step 1: Net shipments: $$N = A - R = \begin{bmatrix}120-5 & 80-2 & 60-1 \\ 90-3 & 50-4 & 70-2\end{bmatrix} = \begin{bmatrix}115 & 78 & 59 \\ 87 & 46 & 68\end{bmatrix}$$ Interpretation: Net shipments represent actual units delivered after returns. Step 2: Unit prices: $$p=\begin{bmatrix}30 \\ 20 \\ 25\end{bmatrix}$$ Step 3: Revenue per customer is matrix multiplication of net shipments and prices: For C1: $$115\times30 + 78\times20 + 59\times25 = 3450 + 1560 + 1475 = 6485$$ For C2: $$87\times30 + 46\times20 + 68\times25 = 2610 + 920 + 1700 = 5230$$ Step 4: Total revenue: $$6485 + 5230 = 11715$$ --- 3. Problem: Calculate total and overall production cost for plants. Production matrix: $$A=\begin{bmatrix}12 & 10 & 8 \\ 15 & 9 & 6\end{bmatrix}$$ Cost per dozen: $$C=\begin{bmatrix}40 \\ 55 \\ 60\end{bmatrix}$$ Step 1: Total cost per plant is matrix multiplication: Plant 1: $$12\times40 + 10\times55 + 8\times60 = 480 + 550 + 480 = 1510$$ Plant 2: $$15\times40 + 9\times55 + 6\times60 = 600 + 495 + 360 = 1455$$ Step 2: Overall cost: $$1510 + 1455 = 2965$$ Step 3: Interpretation: Plant 2 is more efficient as it produces more fans at lower total cost. --- 4. Problem: Compute ending inventory and revenue. Starting inventory: $$S=\begin{bmatrix}80 & 60 & 40 \\ 50 & 70 & 30\end{bmatrix}$$ Units sold: $$D=\begin{bmatrix}65 & 50 & 38 \\ 42 & 58 & 24\end{bmatrix}$$ Prices: $$p=\begin{bmatrix}25 \\ 18 \\ 30\end{bmatrix}$$ Step 1: Ending inventory: $$E = S - D = \begin{bmatrix}80-65 & 60-50 & 40-38 \\ 50-42 & 70-58 & 30-24\end{bmatrix} = \begin{bmatrix}15 & 10 & 2 \\ 8 & 12 & 6\end{bmatrix}$$ Interpretation: Entry E1,3 = 2 means 2 units of product P3 remain in branch B1 after sales. Step 2: Revenue per branch: Branch 1: $$65\times25 + 50\times18 + 38\times30 = 1625 + 900 + 1140 = 3665$$ Branch 2: $$42\times25 + 58\times18 + 24\times30 = 1050 + 1044 + 720 = 2814$$ Step 3: Total revenue: $$3665 + 2814 = 6479$$ Step 4: Product contribution to revenue: P1: $ (65+42)\times25 = 107\times25=2675$ P2: $ (50+58)\times18 = 108\times18=1944$ P3: $ (38+24)\times30 = 62\times30=1860$ P1 contributes most. Recommendation: Increase stock and marketing for P1. --- 5. Problem: Compute total production and revenue, check revenue targets. Planned production: $$B=\begin{bmatrix}180 & 120 & 90 \\ 150 & 100 & 80\end{bmatrix}$$ Overtime: $$O=\begin{bmatrix}20 & 15 & 0 \\ 10 & 0 & 12\end{bmatrix}$$ Prices: $$p=\begin{bmatrix}30 \\ 25 \\ 20\end{bmatrix}$$ Step 1: Total production: $$T = B + O = \begin{bmatrix}200 & 135 & 90 \\ 160 & 100 & 92\end{bmatrix}$$ Step 2: Revenue per plant: Plant A: $$200\times30 + 135\times25 + 90\times20 = 6000 + 3375 + 1800 = 11175$$ Plant B: $$160\times30 + 100\times25 + 92\times20 = 4800 + 2500 + 1840 = 9140$$ Step 3: Total revenue: $$11175 + 9140 = 20315$$ Step 4: Revenue targets: Target = 9500 Plant A exceeds by: $$11175 - 9500 = 1675$$ Plant B shortfall: $$9500 - 9140 = 360$$ --- 6. Problem: Find total cost, total revenue, demand, and consumption functions. Given: $$MC(Q) = 2Q^2 - 6Q + 9$$ Fixed cost = 120 Step 1: Total cost function is integral of marginal cost plus fixed cost: $$TC(Q) = \int MC(Q) dQ + 120 = \int (2Q^2 - 6Q + 9) dQ + 120 = \frac{2}{3}Q^3 - 3Q^2 + 9Q + 120$$ Step 2: Marginal revenue: $$MR(Q) = 80 - 5Q$$ Total revenue is integral: $$TR(Q) = \int MR(Q) dQ = 80Q - \frac{5}{2}Q^2 + C$$ Assuming zero revenue at zero quantity, $C=0$. Step 3: Demand function from $MR = P + Q \frac{dP}{dQ}$, for linear demand: $$MR = P(Q) + Q P'(Q)$$ Given $MR = 80 - 5Q$, demand is: $$P(Q) = 80 - 5Q$$ Step 4: Marginal propensity to consume: $$MPC(Y) = 0.55 + 0.15 Y^{-2/3}$$ Consumption function $C(Y)$ satisfies: $$\frac{dC}{dY} = MPC(Y)$$ Integrate: $$C(Y) = \int (0.55 + 0.15 Y^{-2/3}) dY = 0.55Y + 0.15 \times \frac{Y^{1/3}}{1/3} + K = 0.55Y + 0.45 Y^{1/3} + K$$ Given $C(64) = 100$, and $64^{1/3} = 4$: $$100 = 0.55 \times 64 + 0.45 \times 4 + K = 35.2 + 1.8 + K = 37 + K$$ So, $$K = 63$$ Final consumption function: $$C(Y) = 0.55Y + 0.45 Y^{1/3} + 63$$ --- 7. Problem: Find total cost, total revenue, demand, and consumption functions. Given: $$MC(Q) = Q^2 + 4Q + 6$$ Fixed cost = 200 Step 1: Total cost: $$TC(Q) = \int MC(Q) dQ + 200 = \int (Q^2 + 4Q + 6) dQ + 200 = \frac{1}{3}Q^3 + 2Q^2 + 6Q + 200$$ Step 2: Marginal revenue: $$MR(Q) = 60 - 3Q$$ Total revenue: $$TR(Q) = \int MR(Q) dQ = 60Q - \frac{3}{2}Q^2 + C$$ Assuming $C=0$. Step 3: Demand function: $$P(Q) = 60 - 3Q$$ Step 4: Marginal propensity to consume: $$MPC(Y) = 0.6 + 0.2 Y^{-1/4}$$ Consumption function: $$C(Y) = \int (0.6 + 0.2 Y^{-1/4}) dY = 0.6Y + 0.2 \times \frac{Y^{3/4}}{3/4} + K = 0.6Y + \frac{0.2 \times 4}{3} Y^{3/4} + K = 0.6Y + \frac{0.8}{3} Y^{3/4} + K$$ Given $C(81) = 95$, and $81^{3/4} = (81^{1/4})^3 = 3^3 = 27$: $$95 = 0.6 \times 81 + \frac{0.8}{3} \times 27 + K = 48.6 + 7.2 + K = 55.8 + K$$ So, $$K = 39.2$$ Final consumption function: $$C(Y) = 0.6Y + \frac{0.8}{3} Y^{3/4} + 39.2$$