Let's solve both problems step by step. --- ## Problem 7 **Given:** - Policy A: Premium = $50, $10,000 theft insurance, $180,000 liability insurance. - Policy B: Premium = $40, $15,000 theft insurance, $120,000 liability insurance. - Need at least $300,000 theft and $300,000 liability. - Find how many of each policy (let A = number of Policy A, B = number of Policy B) to minimize premium. ### **Step 1: Define Variables** Let \( x \) = number of Policy A \( y \) = number of Policy B ### **Step 2: Set Up Constraints** #### Theft insurance: \[ 10{,}000x + 15{,}000y \geq 300{,}000 \] #### Liability insurance: \[ 180{,}000x + 120{,}000y \geq 300{,}000 \] #### Non-negativity: \[ x \geq 0,\quad y \geq 0 \] ### **Step 3: Objective Function** Minimize monthly premium: \[ C = 50x + 40y \] --- ### **Step 4: Simplify Constraints** #### Theft: \[ 10,000x + 15,000y \geq 300,000 \implies 2x + 3y \geq 60 \] #### Liability: \[ 180,000x + 120,000y \geq 300,000 \implies 3x + 2y \geq 5 \] But divide by 60,000 to get: \[ 3x + 2y \geq 5 \] But that's not a nice multiple. Let's recalculate: \[ 180,000x + 120,000y \geq 300,000 \implies 18x + 12y \geq 30 \implies 3x + 2y \geq 5 \] So that's correct. #### Final Constraints: \[ 2x + 3y \geq 60 \\ 3x + 2y \geq 5 \\ x \geq 0, \; y \geq 0 \] --- ### **Step 5: Solve System** We want to minimize \( C = 50x + 40y \). The minimum will occur at a vertex of the feasible region. #### Let's solve the two equations as equalities: \[ 2x + 3y = 60 \\ 3x + 2y = 5 \] Multiply the first by 3: \[ 6x + 9y = 180 \] Multiply the second by 2: \[ 6x + 4y = 10 \] Subtract: \[ (6x + 9y) - (6x + 4y) = 180 - 10 \\ 5y = 170 \implies y = 34 \] Plug y into first: \[ 2x + 3(34) = 60 \implies 2x + 102 = 60 \implies 2x = -42 \implies x = -21 \] But x can't be negative. So this intersection is outside our feasible region. #### Try boundary values: ##### 1. Satisfy theft with only Policy B: \[ 10,000x + 15,000y \geq 300,000 \implies x=0: 15,000y \geq 300,000 \implies y \geq 20 \] ##### 2. Satisfy liability with only Policy A: \[ 180,000x + 120,000y \geq 300,000 \implies y=0: 180,000x \geq 300,000 \implies x \geq 1.666... \implies x \geq 1.67 \] But let's check combinations: #### At \( x=0 \): - Theft: \( 15,000y \geq 300,000 \implies y \geq 20 \) - Liability: \( 120,000y \geq 300,000 \implies y \geq 2.5 \implies y \geq 3 \) So lowest integer \( y \geq 20 \). - Cost: \( C = 40(20) = 800 \) #### At \( y=0 \): - Theft: \( 10,000x \geq 300,000 \implies x \geq 30 \) - Liability: \( 180,000x \geq 300,000 \implies x \geq 1.666... \implies x \geq 2 \) So lowest integer \( x \geq 30 \). - Cost: \( C = 50(30) = 1500 \) #### Try intersection of the two constraints as equalities: \[ 2x + 3y = 60 \\ 3x + 2y = 5 \] As shown above, intersection is not feasible. #### Try when one constraint is binding: ##### If \( 2x + 3y = 60 \): \( 3x + 2y \geq 5 \) Let’s solve for x: \[ 2x + 3y = 60 \implies x = (60 - 3y)/2 \] Plug into \( 3x + 2y \geq 5 \): \[ 3\left(\frac{60 - 3y}{2}\right) + 2y \geq 5 \\ \frac{180 - 9y + 4y}{2} \geq 5 \\ \frac{180 - 5y}{2} \geq 5 \\ 180 - 5y \geq 10 \implies 5y \leq 170 \implies y \leq 34 \] But since \( x \geq 0 \), \( (60 - 3y)/2 \geq 0 \implies 60 - 3y \geq 0 \implies y \leq 20 \) So \( 0 \leq y \leq 20 \) So try \( y = 20 \): - \( x = (60 - 3(20))/2 = (60 - 60)/2 = 0 \) - Check other constraint: \( 3(0) + 2(20) = 40 \geq 5 \) OK - Cost: \( C = 50(0) + 40(20) = 800 \) Try \( y = 0 \): - \( x = 60/2 = 30 \) - Check other constraint: \( 3(30) + 2(0) = 90 \geq 5 \) OK - Cost: \( C = 50(30) = 1500 \) Now, try integer values between 0 and 20 for \( y \): Let’s see if lower cost is possible by mixing policies. Try \( y = 10 \): - \( x = (60 - 3(10))/2 = (60 - 30)/2 = 15 \) - Check \( 3x + 2y = 3(15) + 2(10) = 45 + 20 = 65 \geq 5 \) OK - Cost: \( 50(15) + 40(10) = 750 + 400 = 1150 \) So, the minimum is at \( x = 0, y = 20 \), with a cost of $800. --- ### **Final Answer for Problem 7** **Purchase 0 units of Policy A and 20 units of Policy B. The minimum monthly premium is $800.** --- ## Problem 8 **Given:** - Two types of phone cases: Case 1 and Case 2. - Each case uses time on 3 machines. - Machine 1: 100 minutes, Machine 2: 720 minutes, Machine 3: 300 minutes. - Case 1 uses: 0.4 (M1), 0.6 (M2), 0.2 (M3) - Case 2 uses: 0.8 (M1), 0.2 (M2), 0.2 (M3) - Case 1 sells for $15, Case 2 for $20. - Let \( x \) = # of Case 1, \( y \) = # of Case 2. ### **Step 1: Set Up Constraints** \[ \begin{align*} \text{M1:} & \quad 0.4x + 0.8y \leq 100 \\ \text{M2:} & \quad 0.6x + 0.2y \leq 720 \\ \text{M3:} & \quad 0.2x + 0.2y \leq 300 \\ x \geq 0,\, y \geq 0 \end{align*} \] ### **Step 2: Objective Function** \[ \text{Maximize } R = 15x + 20y \] --- ### **Step 3: Simplify Constraints** #### M3: \[ 0.2x + 0.2y \leq 300 \implies x + y \leq 1500 \] #### M2: \[ 0.6x + 0.2y \leq 720 \implies 3x + y \leq 3600 \] #### M1: \[ 0.4x + 0.8y \leq 100 \implies x + 2y \leq 250 \] --- ### **Step 4: Find Vertices (Feasible Region)** We'll solve intersections of: 1. \( x + y \leq 1500 \) 2. \( 3x + y \leq 3600 \) 3. \( x + 2y \leq 250 \) #### (a) **Intersection of (1) and (3):** \( x + y = 1500 \) \( x + 2y = 250 \) Subtract equations: \( (x + 2y) - (x + y) = 250 - 1500 \implies y = -1250 \) Plug \( y = -1250 \) into \( x + y = 1500 \implies x = 2750 \) But \( y \geq 0 \implies \) not feasible. #### (b) **Intersection of (1) and (2):** \( x + y = 1500 \) \( 3x + y = 3600 \) Subtract equations: \( (3x + y) - (x + y) = 3600 - 1500 \implies 2x = 2100 \implies x = 1050 \) Then \( y = 1500 - 1050 = 450 \) Check (3): \( x + 2y = 1050 + 2(450) = 1950 \leq 250? \) **No**! Not feasible. #### (c) **Intersection of (2) and (3):** \( 3x + y = 3600 \) \( x + 2y = 250 \) Let’s solve: From (3): \( x = 250 - 2y \) Plug into (2): \( 3(250 - 2y) + y = 3600 \implies 750 - 6y + y = 3600 \implies 750 - 5y = 3600 \implies -5y = 2850 \implies y = -570 \) Plug back: \( x = 250 - 2(-570) = 250 + 1140 = 1390 \) But \( y \geq 0 \implies \) not feasible. #### (d) **Intercepts** Try \( y = 0 \): - (1): \( x \leq 1500 \) - (2): \( 3x \leq 3600 \implies x \leq 1200 \) - (3): \( x \leq 250 \) So, **maximum \( x = 250 \) at \( y = 0 \)** Revenue: \( R = 15(250) + 20(0) = 3750 \) Try \( x = 0 \): - (1): \( y \leq 1500 \) - (2): \( y \leq 3600 \) - (3): \( 2y \leq 250 \implies y \leq 125 \) So, **maximum \( y = 125 \) at \( x = 0 \)** Revenue: \( R = 15(0) + 20(125) = 2500 \) #### Try when \( x + 2y = 250 \): Let’s try maximizing \( R \) along this boundary. From \( x + 2y = 250 \implies x = 250 - 2y \) Plug into other constraints: (1): \( x + y = (250 - 2y) + y = 250 - y \leq 1500 \implies y \geq -1250 \) (always true for \( y \geq 0 \)) (2): \( 3x + y = 3(250 - 2y) + y = 750 - 6y + y = 750 - 5