The Ride This skate board factory produces Mega skateboards and Supreme skateboards. their maximum output is 40 skateboards a day and they must produce at least 3 times as many Mega boards as Supreme boards. the profit on a Mega skateboard is $55 and the profit on a Supreme skateboard is $75. What is the factory's maximum daily profit?

To solve this problem, we need to set up a system of inequalities and an objective function to maximize the daily profit of the skateboard factory.

**Step 1: Define Variables**
- Let $ M $ represent the number of Mega skateboards produced daily.
- Let $ S $ represent the number of Supreme skateboards produced daily.

**Step 2: Write the Constraints**
1. The total number of skateboards produced cannot exceed 40:
   \[
   M + S \leq 40
   \]

2. The factory must produce at least 3 times as many Mega skateboards as Supreme skateboards:
   \[
   M \geq 3S
   \]

3. Both $ M $ and $ S $ must be non-negative:
   \[
   M \geq 0 \quad \text{and} \quad S \geq 0
   \]

**Step 3: Define the Objective Function**
- The profit from Mega skateboards is $55 per board, so the profit from $ M $ Mega skateboards is $ 55M $.
- The profit from Supreme skateboards is $75 per board, so the profit from $ S $ Supreme skateboards is $ 75S $.
  
The objective function (total profit $ P $) to maximize is:
\[
P = 55M + 75S
\]

**Step 4: Solve the System of Inequalities**
We need to find the values of $ M $ and $ S $ that maximize $ P = 55M + 75S $ under the constraints.

**Step 5: Check the Vertices of the Feasible Region**
The feasible region is determined by the constraints:
1. $ M + S = 40 $
2. $ M = 3S $

Now, let's find the intersection points:

1. Set $ M = 3S $ into $ M + S = 40 $:
   \[
   3S + S = 40 \implies 4S = 40 \implies S = 10
   \]
   Substituting $ S = 10 $ into $ M = 3S $:
   \[
   M = 3 \times 10 = 30
   \]
   So, one vertex is $ (M, S) = (30, 10) $.

2. Other vertices are found by testing when $ S = 0 $:
   - If $ S = 0 $, then $ M + 0 = 40 $ implies $ M = 40 $.
   - If $ M = 0 $, then $ S = 0 $ because $ M = 3S $.

**Step 6: Evaluate the Profit at Each Vertex**
1. At $ (M, S) = (30, 10) $:
   \[
   P = 55 \times 30 + 75 \times 10 = 1650 + 750 = 2400
   \]

2. At $ (M, S) = (40, 0) $:
   \[
   P = 55 \times 40 + 75 \times 0 = 2200
   \]

3. At $ (M, S) = (0, 0) $:
   \[
   P = 55 \times 0 + 75 \times 0 = 0
   \]

**Step 7: Conclusion**
The maximum profit occurs at $ (M, S) = (30, 10) $, with a maximum daily profit of:

\[
\boxed{2400}
\]

Would you like more details on any part of this solution, or do you have any questions?

Here are some related questions for further exploration:

1. How would the maximum profit change if the profit per Mega skateboard increased to $60?
2. What would happen to the maximum profit if the constraint $ M \geq 3S $ changed to $ M \geq 2S $?
3. How would

The Ride This skate board factory produces Mega skateboards and Supreme skateboards. Their maximum output is 40 skateboards a day and they must produce at least 3 times as many Mega boards as Supreme boards. The profit on a Mega skateboard is $55 and the profit on a Supreme skateboard is $75. What is the factory's maximum daily profit?

Learn how to maximize the daily profit for a skateboard factory producing Mega and Supreme skateboards using linear programming. The factory must produce at least three times as many Mega boards as Supreme boards, with a maximum output of 40 boards per day. This problem covers optimization, inequality constraints, and profit calculations. Suitable for students in Grades 10-12.

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