# Furniture Company Linear Programming Problem Let's define variables: - Let \( x \) = number of desks produced - Let \( y \) = number of chairs produced ## Step 1: Define the Problem ### Objective Maximize profit: \[ \text{Profit} = 400x + 250y \] ### Constraints 1. **Wood constraint:** Each desk uses 4 units, each chair uses 3 units, and there are 200 units available: \[ 4x + 3y \leq 200 \] 2. **Marketing constraint:** The number of chairs must be at least twice the number of desks: \[ y \geq 2x \] 3. **Non-negativity:** \[ x \geq , \quad y \geq \] --- ## Step 2: Set Up the Problem for Solver ### Write Constraints Explicitly 1. \( 4x + 3y \leq 200 \) 2. \( y - 2x \geq \) 3. \( x \geq \) 4. \( y \geq \) --- ## Step 3: Solve Graphically/Algebraically ### Find Intersection Points #### 1. Where \( y = 2x \) and \( 4x + 3y = 200 \): Substitute \( y = 2x \) into the wood constraint: \[ 4x + 3(2x) = 200 \implies 4x + 6x = 200 \implies 10x = 200 \implies x = 200 \] \[ y = 2x = 400 \] #### 2. Where \( x = \): Plug \( x = \) into \( 4x + 3y \leq 200 \): \[ 3y \leq 200 \implies y \leq 666.67 \] But from marketing, \( y \geq 2x = \), so \( y \geq \). - Corner point: \( (, 666.67) \) #### 3. Where \( y = \): Plug \( y = \) into \( 4x + 3y \leq 200 \): \[ 4x \leq 200 \implies x \leq 500 \] But marketing: \( \geq 2x \implies x \leq \). So only feasible point is \( x = , y = \). --- ### Calculate Profit at Feasible Points - \( (, ) \): \( 400* + 250* = \$ \) - \( (, 666.67) \): \( 400* + 250*666.67 = \$166,667.50 \) - \( (200, 400) \): \( 400*200 + 250*400 = 80,000 + 100,000 = \$180,000 \) --- ## Step 4: Check Feasibility - For \( (, 666.67) \): - \( 4* + 3*666.67 = 200 \): OK - \( y \geq 2x \rightarrow 666.67 \geq \): OK - For \( (200, 400) \): - \( 4*200 + 3*400 = 800 + 120 = 200 \): OK - \( 400 \geq 2*200 = 400 \): OK - For \( (,) \): trivial. --- ## Step 5: Final Answer for (a) **Maximum Profit:** \[ \boxed{\$180,000} \] **Produce:** - \( 200 \) desks - \( 400 \) chairs --- # (b) Sensitivity Analysis Using SolverTable ## Varying Wood Availability from 100 to 300 (in 100-unit increments) Let \( W \) be the available wood units. ### Steps: 1. For each value of \( W \), solve: - \( 4x + 3y \leq W \) - \( y \geq 2x \) - \( x, y \geq \) - Maximize \( 400x + 250y \) ### Key points: - When the optimal solution is at \( y = 2x \), substitute into wood constraint: \[ 4x + 3(2x) = 4x + 6x = 10x \leq W \implies x \leq W/10, y = 2x \] - Profit: \( 400x + 250y = 400x + 250*2x = 900x \) - When the optimal solution is at \( x= \), then \( y \leq W/3 \), profit: \( 250y \) ### Compare which profit is higher for each \( W \): #### Profit at \( x=, y=W/3 \): \( 250(W/3) = (250/3)W \) #### Profit at \( x=W/10, y=2W/10 \): \( 900(W/10) = 90W \) Set equal to find the switching point: \[ 90W = (250/3)W \implies 90 = 83.33 \] So, for any \( W \), \( 90W > 83.33W \), so solution at \( y=2x \) will always be better (unless constrained by non-negativity). #### Check if solution is feasible for all \( W \geq \): - \( x = W/10 \geq \) always. - \( y = 2W/10 = W/5 \geq \) ### Table of Results | Wood \( W \) | \( x = W/10 \) | \( y = W/5 \) | Profit \( 900x \) | |:------------:|:--------------:|:-------------:|:-----------------:| | 100 | 100 | 200 | 90,000 | | 110 | 110 | 220 | 99,000 | | 120 | 120 | 240 | 108,000 | | 130 | 130 | 260 | 117,000 | | 140 | 140 | 280 | 126,000 | | 150 | 150 | 300 | 135,000 | | 160 | 160 | 320 | 144,000 | | 170 | 170 | 340 | 153,000 | | 180 | 180 | 360 | 162,000 | | 190 | 190 | 380 | 171,000 | | 200 | 200 | 400 | 180,000 | | 210 | 210 | 420 | 189,000 | | 220 | 220 | 440 | 198,000 | | 230 | 230 | 460 | 207,000 | | 240 | 240 | 480 | 216,000 | | 250 | 250 | 500 | 225,000 | | 260 | 260 | 520 | 234,000 | | 270 | 270 | 540 | 243,000 | | 280 | 280 | 560 | 252,000 | | 290 | 290 | 580 | 261,000 | | 300 | 300 | 600 | 270,000 | --- ## Marginal Profit per Extra Unit of Wood - From the table, as \( W \) increases by 100, profit increases by \( 900 \times 10 = 9,000 \). - So per unit increase in \( W \): \[ \text{Marginal Profit} = \frac{9,000}{100} = \$90 \] - This is the **shadow price** of the wood constraint. **Interpretation:** - The company would be willing to pay up to **\$90 for each extra unit of wood** above 200 units (and would lose \$90 for every unit lost, as long as the shadow price stays valid). --- ## Summary Table | Wood Available | Desks Produced | Chairs Produced | Profit (\$) | |:--------------:|:--------------:|:--------------:|:-----------:| | 100 | 100 | 200 | 90,000 | | ... | ... | ... | ... | | 200 | 200 | 400 | 180,000 | | ... | ... | ... | ... | | 300 | 300 | 600 | 270,000 | --- ## Final Answers ### (a) Maximum Profit \[ \boxed{ \text{Produce 200 desks and 400 chairs for a maximum profit of \$180{,}000.} } \] ### (b) Sensitivity Analysis - **For each extra unit of wood above 200 units, profit increases by \$90.** - **For each unit of wood lost below 200 units, profit decreases by \$90.** **How much to pay for extra wood?** Up to \$90 per unit. **How much profit is lost if any of the 200 units are lost?** \$90 per unit lost. --- **Note:** This shadow price (\$90) is valid as long as the optimal solution remains at \( y = 2x \) and the constraints do not change structure (which happens for all values within the table above).

# Furniture Company Linear Programming Problem Let's define variables: - Let \( x \) = number of desks produced - Let \( y \) = number of chairs produced ## Step 1: Define the Problem ### Objective Maximize profit: \[ \text{Profit} = 400x + 250y \] **Explanation:** This equation represents the total profit generated from selling desks and chairs. The coefficients (400 for desks and 250 for chairs) indicate the profit per unit sold. ### Constraints 1. **Wood constraint:** Each desk uses 4 units of wood, each chair uses 3 units, and there are 200 units available: \[ 4x + 3y \leq 200 \] **Explanation:** This inequality ensures that the total amount of wood used for desks and chairs does not exceed the available wood. 2. **Marketing constraint:** The number of chairs must be at least twice the number of desks: \[ y \geq 2x \] **Explanation:** This constraint reflects marketing requirements that dictate the ratio of chairs to desks. 3. **Non-negativity constraints:** \[ x \geq 0, \quad y \geq 0 \] **Explanation:** These constraints ensure that the number of desks and chairs produced cannot be negative. --- ## Step 2: Set Up the Problem for Solver ### Write Constraints Explicitly 1. \( 4x + 3y \leq 200 \) 2. \( y - 2x \geq 0 \) 3. \( x \geq 0 \) 4. \( y \geq 0 \) **Explanation:** This step formalizes the constraints in a way that can be input into a solver, which will handle the optimization. --- ## Step 3: Solve Graphically/Algebraically ### Find Intersection Points #### 1. Where \( y = 2x \) and \( 4x + 3y = 200 \): Substitute \( y = 2x \) into the wood constraint: \[ 4x + 3(2x) = 200 \implies 4x + 6x = 200 \implies 10x = 200 \implies x = 20 \] \[ y = 2x = 40 \] **Explanation:** This calculation finds the intersection point of the marketing constraint and the wood constraint, giving a potential solution. #### 2. Where \( x = 0 \): Plug \( x = 0 \) into \( 4x + 3y \leq 200 \): \[ 3y \leq 200 \implies y \leq 66.67 \] Further, from marketing, \( y \geq 0 \). So the feasible point is \( (0, 66.67) \). **Explanation:** This scenario checks the case where no desks are produced and maximizes chair production under the wood constraint. #### 3. Where \( y = 0 \): Plug \( y = 0 \) into \( 4x + 3y \leq 200 \): \[ 4x \leq 200 \implies x \leq 50 \] So the feasible point is \( (50, 0) \). **Explanation:** This checks the case where only desks are produced, maximizing production within the wood constraint. ### Calculate Profit at Feasible Points - \( (0, 66.67) \): Profit = \( 400*0 + 250*66.67 = 16,667.50 \) - \( (20, 40) \): Profit = \( 400*20 + 250*40 = 8,000 + 10,000 = 18,000 \) - \( (50, 0) \): Profit = \( 400*50 + 250*0 = 20,000 \) **Explanation:** This step calculates the profit for each feasible production combination using the profit formula. --- ## Step 4: Check Feasibility - For \( (0, 66.67) \): - \( 4*0 + 3*66.67 = 200 \): OK - \( 66.67 \geq 2*0 = 0 \): OK - For \( (20, 40) \): - \( 4*20 + 3*40 = 80 + 120 = 200 \): OK - \( 40 \geq 2*20 = 40 \): OK - For \( (50, 0) \): - \( 4*50 + 3*0 = 200 \): OK - \( 0 \geq 2*50 = 100 \): Not OK **Explanation:** This step verifies that the calculated points meet all constraints, ensuring they are valid solutions. --- ## Step 5: Final Answer for (a) **Maximum Profit:** \[ \boxed{\$20,000} \] **Produce:** - \( 50 \) desks - \( 0 \) chairs --- # (b) Sensitivity Analysis Using SolverTable ## Varying Wood Availability from 100 to 300 (in 100-unit increments) Let \( W \) be the available wood units. ### Steps: 1. For each value of \( W \), solve: - \( 4x + 3y \leq W \) - \( y \geq 2x \) - \( x, y \geq 0 \) - Maximize \( 400x + 250y \) **Explanation:** This step outlines how to systematically evaluate the impact of varying wood availability on production and profit. ### Key Points: - When the optimal solution is at \( y = 2x \), substitute into the wood constraint: \[ 4x + 3(2x) = 4x + 6x = 10x \leq W \implies x \leq \frac{W}{10}, \, y = 2x \] - Profit: \( 400x + 250y = 400x + 250(2x) = 900x \) - When the optimal solution is at \( x = 0 \), then \( y \leq \frac{W}{3} \), profit: \( 250y \) **Explanation:** This breakdown compares profit outcomes based on the optimal solutions depending on constraints. ### Compare Profits for Each \( W \): #### Profit at \( x=\frac{W}{10}, y=\frac{2W}{10} \): Profit = \( 900 \left(\frac{W}{10}\right) = 90W \) #### Profit at \( y=\frac{W}{3} \): Profit = \( 250 \left(\frac{W}{3}\right) = \frac{250}{3}W \) Set equal to find the switching point: \[ 90W = \frac{250}{3}W \implies 90 = 83.33 \] **Explanation:** This comparison determines when profit from producing chairs only becomes equal to profit from producing desks and chairs together. ### Table of Results | Wood \( W \) | \( x = \frac{W}{10} \) | \( y = \frac{W}{5} \) | Profit \( 900x \) | |:------------:|:----------------------:|:---------------------:|:-----------------:| | 100 | 10 | 20 | 9,000 | | 200 | 20 | 40 | 18,000 | | 300 | 30 | 60 | 27,000 | --- ## Marginal Profit per Extra Unit of Wood - From the table, as \( W \) increases by 100, profit increases by \( 900 \times 10 = 9,000 \). - So per unit increase in \( W \): \[ \text{Marginal Profit} = \frac{9,000}{100} = \$90 \] - This is the **shadow price** of the wood constraint. **Interpretation:** - The company would be willing to pay up to **\$90 for each extra unit of wood** above 200 units (and would lose \$90 for every unit lost, as long as the shadow price stays valid). --- ## Summary Table | Wood Available | Desks Produced | Chairs Produced | Profit (\$) | |:--------------:|:--------------:|:--------------:|:-----------:| | 100 | 10 | 20 | 9,000 | | 200 | 20 | 40 | 18,000 | | 300 | 30 | 60 | 27,000 | --- ## Final Answers ### (a) Maximum Profit \[ \boxed{ \text{Produce 20 desks and 40 chairs for a maximum profit of \$18,000.} } \] ### (b) Sensitivity Analysis - **For each extra unit of wood above 200 units, profit increases by \$90.** - **For each unit of wood lost below 200 units, profit decreases by \$90.** **How much to pay for extra wood?** Up to \$90 per unit. **How much profit is lost if any of the 200 units are lost?** \$90 per unit lost. --- **Note:** This shadow price (\$90) is valid as long as the optimal solution remains at \( y = 2x \) and the constraints do not change structure (which happens for all values within the table above).