Define Variables: First, let's define the variables: $\newline$ Let $L$ be the number of LENCIO cars produced per week. $\newline$ 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: $\newline$ $L + S \leq 160$
Second 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: $\newline$ $20L + 12S \leq 2400$
Third Constraint: Each LENCIO uses $550\,\text{kg}$ and a STEM uses $1100\,\text{kg}$ of components. The factory has $154,000\,\text{kg}$ of components available. This gives us our third constraint: $\newline$ $550L + 1100S \leq 154,000$
Fourth Constraint: They have to make at least $50$ STEMS a week to meet an existing contract. This gives us our fourth constraint: $\newline$ $S \geq 50$
Objective 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: $\newline$ Profit = $8700L + 6800S$
Solve 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. $\newline$ 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. $\newline$ From $L + S = 160$ , we get $L = 160 - S$ . $\newline$ Substitute $L$ in the first equation: $20(160 - S) + 12S = 2400$ .
Intersection Point $2$ : Solving for $S$ , we get: $\newline$ $3200 - 20S + 12S = 2400$ $\newline$ $8S = 800$ $\newline$ $S = 100$ $\newline$ Then, $L = 160 - S = 160 - 100 = 60$ $\newline$ 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: $\newline$ $88,000 - 550S + 1100S = 154,000$ $\newline$ $550S = 66,000$ $\newline$ $S = 120$ $\newline$ Then, $L = 160 - S = 160 - 120 = 40$ $\newline$ 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: $\newline$ For $(60, 100)$ : Profit $= 8700\times60 + 6800\times100 = 522,000 + 680,000 = 1,202,000$ $\newline$ For $(40, 120)$ : Profit $= 8700\times40 + 6800\times120 = 348,000 + 816,000 = 1,164,000$ $\newline$ For $(0, 160)$ : Profit $= 8700\times0 + 6800\times160 = 0 + 1,088,000 = 1,088,000$ $\newline$ For $(160, 0)$ : Profit $= 8700\times160 + 6800\times0 = 1,392,000 + 0 = 1,392,000$ $\newline$ 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: $\newline$ $20L + 12\times50 \leq 2400$ $\newline$ $20L + 600 \leq 2400$ $\newline$ $20L \leq 1800$ $\newline$ $L \leq 90$ $\newline$ So the maximum $L$ we can produce is $90$ when $S = 50$ .
Calculate Profit - Point $5$ : Now we calculate the profit for $(90, 50)$ : $\newline$ Profit $= 8700\times90 + 6800\times50 = 783,000 + 340,000 = 1,123,000$
Find Maximum L: Comparing all the profits, we see that the maximum profit is when $(L, S) = (60, 100)$ , which gives a profit of $\$1,202,000$ .