A space shuttle orbits Earth one time in 90 minutes. How many times does orbit earth in six hours?
2 answers:
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Answer:
- The solution that optimizes the profit is producing 0 small lifts and 50 large lifts.
- Below are all the steps explained in detail.
- The graph is attached.
Explanation:
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<u>1. Name the variables:</u>
- x: number of smaller lifts
- y: number of larger lifts
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<u>2. Build a table to determine the number of hours each lift requires from each department:</u>
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Number of hours
small lift large lift total per department
Welding department 1x 3y x + 3y
Packaging department 2x 1y 2x + y
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<u>3. Constraints</u>
- 150 hours available in welding: x + 3y ≤ 150
- 120 hours available in packaging: 2x + y ≤ 120
- The variables cannot be negative: x ≥ 0, and y ≥ 0
Then you must:
- draw the lines and regions defined by each constraint
- determine the region of solution that satisfies all the constraints
- determine the vertices of the solution region
- test the profit function for each of the vertices. The vertex that gives the greatest profit is the solution (the number of each tupe that should be produced to maximize profits)
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<u>4. Graph</u>
See the graph attached.
Here is how you draw it.
- x + 3y ≤ 150
- draw the line x + 3y = 150 (a solid line because it is included in the solution set)
- shade the region below and to the left of the line
- 2x + y ≤ 120
- draw the line 2x + y ≤ 120 (a solid line because it is included in the solution set)
- shade the region below and to the left of the line
- x ≥ 0 and y ≥ 0: means that only the first quadrant is considered
- the solution region is the intersection of the regions described above.
- take the points that are vertices inside the solutoin region.
<u>5. Test the profit function for each vertex</u>
The profit function is P(x,y) = 25x + 90y
The vertices shown in the graph are:
- (0,0)
- (0,50)
- (42,36)
- (60,0)
The profits with the vertices are:
- P(0,0) = 0
- P(0,50) = 25(0) + 90(50) = 4,500
- P(42,36) = 25(42) + 90(36) = 4,290
- P(60,0) = 25(60) + 90(0) = 1,500
Thus, the solution that optimizes the profit is producing 0 smaller lifts and 90 larger lifts.
