User The car factory can make a maximum of 160 vehicles per week. It takes 20 hours of labour to make a LENCIO and 12 hours for a STEM. Each LENCIO uses 550kg and a STEM uses 1100kg of components to make. The factory has 2400 hours of labour and 154,000 kilograms of components available to use. They have to make at least 50 STEMS a week to meet an existing contract. The factory makes a profit of $8700 for every LENCIO and, $6,800 for every STEM they sell. How many LENCIO and STEM cars do Zapp Automotive need to make each week to maximise their profit?

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Define Variables: First, let's define the variables: Let L be the number of LENCIO cars produced per week. Let S be the number of STEM cars produced per week.First Constraint: The factory can make a maximum of 160 vehicles per week. This gives us our first constraint: L + S ≤ 160Second Constraint: It takes 20 hours of labor to make a LENCIO and 12 hours for a STEM. The factory has 2400 hours of labor available. This gives us our second constraint: 20L + 12S ≤ 2400Third Constraint: Each LENCIO uses 550kg and a STEM uses 1100kg of components. The factory has 154,000 kilograms of components available. This gives us our third constraint: 550L + 1100S ≤ 154,000Fourth Constraint: They have to make at least 50 STEMS a week to meet an existing contract. This gives us our fourth constraint: S ≥ 50Objective Function: The factory makes a profit of $8700 for every LENCIO and $6,800 for every STEM they sell. We want to maximize the profit, so our objective function to maximize is: Profit = 8700L + 6800SSolve System of Inequalities: Now we need to solve the system of inequalities to find the number of LENCIO (L) and STEM (S) cars that maximize the profit. This is a linear programming problem that can be solved using methods such as graphing the feasible region, using the Simplex method, or computational tools. However, since we are not using computational tools here, we will check the corner points of the feasible region determined by the constraints to find the maximum profit.Find Corner Points: First, we need to find the corner points of the feasible region. We do this by solving the system of equations obtained from the constraints. We will find the intersection points of the lines formed by the constraints.Intersection Point 1: Let's find the intersection of 20L + 12S = 2400 and L + S = 160 by solving these two equations simultaneously. From L + S = 160, we get L = 160 - S. Substitute L in the first equation: 20(160 - S) + 12S = 2400.Intersection Point 2: Solving for S, we get: 3200 - 20S + 12S = 2400 8S = 800 S = 100 Then, L = 160 - S = 160 - 100 = 60 So one corner point is (L, S) = (60, 100).Consider Additional Points: Next, let's find the intersection of 550L + 1100S = 154,000 and L + S = 160. From L + S = 160, we get L = 160 - S. Substitute L in the first equation: 550(160 - S) + 1100S = 154,000.Calculate Profit - Point 1: Solving for S, we get: 88,000 - 550S + 1100S = 154,000 550S = 66,000 S = 120 Then, L = 160 - S = 160 - 120 = 40 So another corner point is (L, S) = (40, 120).Calculate Profit - Point 2: We also need to consider the points (L, S) = (0, 160) and (L, S) = (160, 0), which are the intersections of each constraint with the axes, and the point (L, S) = (L, 50) which is the minimum number of STEM cars that need to be produced.Calculate Profit - Point 3: Now we calculate the profit for each of these corner points: For (60, 100): Profit = 8700*60 + 6800*100 = 522,000 + 680,000 = 1,202,000 For (40, 120): Profit = 8700*40 + 6800*120 = 348,000 + 816,000 = 1,164,000 For (0, 160): Profit = 8700*0 + 6800*160 = 0 + 1,088,000 = 1,088,000 For (160, 0): Profit = 8700*160 + 6800*0 = 1,392,000 + 0 = 1,392,000 For (L, 50): We need to find the maximum L that can be produced with the remaining constraints.Calculate Profit - Point 4: To find the maximum L when S = 50, we use the labor constraint: 20L + 12*50 ≤ 2400 20L + 600 ≤ 2400 20L ≤ 1800 L ≤ 90 So the maximum L we can produce is 90 when S = 50.Calculate Profit - Point 5: Now we calculate the profit for (90, 50): Profit = 8700*90 + 6800*50 = 783,000 + 340,000 = 1,123,000Find Maximum L: Comparing all the profits, we see that the maximum profit is when (L, S) = (60, 100), which gives a profit of $1,202,000.

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