A company manufactures two kinds of ice-cream. The vanilla ice-cream sells for $2.50 each while the chocolate flavour ice-cream sells for $4.50 cents each. It costs the company 1 labour hour to make the vanilla flavour ice-cream and 2 labour hours to make the chocolate flavour ice-cream. The company has a total of 300 labour hours available. It costs the company 3 machine hours for the vanilla ice-cream and 2 machine hours for the chocolate ice-cream. The company has a total of 480 machine hours available. How much of each type of ice-cream should the company produce to maximise revenue? What is the maximum revenue?
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- Thread starter Ashna
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Please read the rules for the thread:
You should submit some work for the problem, and we can check it, and give you ideas about what to do next.
In your exercise, you should begin by selecting variables for the unknown quantties in the problem (usually mentioned in the question at the end of the problem). Then write an inequality for labor hours, an inequality for machine hours, and an equation for revenue.
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You should submit some work for the problem, and we can check it, and give you ideas about what to do next.
In your exercise, you should begin by selecting variables for the unknown quantties in the problem (usually mentioned in the question at the end of the problem). Then write an inequality for labor hours, an inequality for machine hours, and an equation for revenue.
| Ice cream | x = Vanilla | y = Chocolate | |
| Labour | 1 | 2 | ≤300 |
| Machine | 3 | 2 | ≤480 |
| S=2.5x+4.5y |
Maximize revenue:
Subject to:
x+2y <= 300
3x + 2y <=480
x >= 0
y >= 0
| Vertex | Value | |
| (90,105) | x+2y=300 3x+2y=480 | 697.50 |
| (0,150) | x+2y=300 x=0 | 675 |
| (160,0) | 3x+2y=480 y=0 | 400 |
| (0,0) | x=0 y=0 | 0 |
All your work is correct. Good job! In your table, you labeled x = vanilla and y = chocolate. That is not quite right. x and y must be numbers so
x = servings of vanilla and y = servings of chocolate. Use this to finish the problem. Your chart has all the information you need. Now tell how many servings of vanilla and how many servings of chocolate will maximize revenue. That will be the final answer for the problem.
x = servings of vanilla and y = servings of chocolate. Use this to finish the problem. Your chart has all the information you need. Now tell how many servings of vanilla and how many servings of chocolate will maximize revenue. That will be the final answer for the problem.
Malia Losalu
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Constraints:
x + 2y <= 300 ; 3x + 2y <= 480 ; x >= 0 ; y >= 0
For the first constraint - can you plot:
x + 2y - 300 = 0
The area UNDER the plotted line (including the line) will be the "constraint".
Similarly, For the second constraint - you plot:
3x + 2y - 480 = 0
The area UNDER the plotted line (including the line) will be the "constraint".
The first quadrant (including the axes), will define the third and the fourth constraints.
x + 2y <= 300 ; 3x + 2y <= 480 ; x >= 0 ; y >= 0
For the first constraint - can you plot:
x + 2y - 300 = 0
The area UNDER the plotted line (including the line) will be the "constraint".
Similarly, For the second constraint - you plot:
3x + 2y - 480 = 0
The area UNDER the plotted line (including the line) will be the "constraint".
The first quadrant (including the axes), will define the third and the fourth constraints.
Malia Losalu
New member
- Joined
- May 3, 2020
- Messages
- 4
- Joined
- Apr 12, 2005
- Messages
- 11,339
"x + 2y <= 300 ; 3x + 2y <= 480 ; x >= 0 ; y >= 0"
I never liked the terms "under" or "over" for this sort of thing. If your boundary gets anywhere near vertical, or if your boundary is not linear, it just isn't always super clear which side to shade. I've always found it easier just to pick a point and see!
x + 2y <= 300
Try (0,0)
0 + 2(0) = 0 <= 300. Perfect. Shade the side that includes (0,0).
3x + 2y <= 480
Try (100,100)
3(100) + 2(100) = 500 > 480. Whoops!. Shade the side that DOES NOT include (100,100).
Just pick a point. (0,0) is often pretty easy. There may be more convenient options.
I never liked the terms "under" or "over" for this sort of thing. If your boundary gets anywhere near vertical, or if your boundary is not linear, it just isn't always super clear which side to shade. I've always found it easier just to pick a point and see!
x + 2y <= 300
Try (0,0)
0 + 2(0) = 0 <= 300. Perfect. Shade the side that includes (0,0).
3x + 2y <= 480
Try (100,100)
3(100) + 2(100) = 500 > 480. Whoops!. Shade the side that DOES NOT include (100,100).
Just pick a point. (0,0) is often pretty easy. There may be more convenient options.