Question
In the preparation of a food supplement of high nutritional value, 8 pounds are available of component A to prepare three presentations (portions) of different sizes (1, 2, and 3). Size 1 contains 50 grams, size 2, 40 grams, and size 3, 20 grams. Are needed at least 5 servings of size 1, at least half of size 2 compared to size 1, and at least twice as many servings of the smallest as of the largest. The portion of Largest portion generates a profit of $23.00, middle portion generates a profit of $34.00 and the smallest portion generates a profit of $38.00. How many servings should be made? of each class so that the benefit is maximum?
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Answer to a math question In the preparation of a food supplement of high nutritional value, 8 pounds are available of component A to prepare three presentations (portions) of different sizes (1, 2, and 3). Size 1 contains 50 grams, size 2, 40 grams, and size 3, 20 grams. Are needed at least 5 servings of size 1, at least half of size 2 compared to size 1, and at least twice as many servings of the smallest as of the largest. The portion of Largest portion generates a profit of $23.00, middle portion generates a profit of $34.00 and the smallest portion generates a profit of $38.00. How many servings should be made? of each class so that the benefit is maximum?
Madelyn
4.768 Answers
1. Convert the weight constraint to grams:
50x + 40y + 20z \leq 3630.739
2. Express constraints in terms of inequalities:
x \geq 5
y \geq \frac{x}{2}
z \geq 2x
3. Substitute the minimum values for y and z based on x:
y = \frac{x}{2}
z = 2x
4. Replace y and z in the weight constraint:
50x + 40 \left(\frac{x}{2}\right) + 20(2x) \leq 3630.739
50x + 20x + 40x \leq 3630.739
110x \leq 3630.739
x \leq \frac{3630.739}{110}
x \leq 33
5. Calculate the maximum total profit:
- For \( x = 5 \):
y = \frac{5}{2} = 2.5
z = 2 \times 5 = 10
Total Profit:
P = 23(5) + 34(3) + 38(10) = 115 + 102 + 380 = 597
- For \( x = 33 \):
y = \frac{33}{2} = 16.5
z = 2 \times 33 = 66
Total Profit:
P = 23(33) + 34(16) + 38(66) = 759 + 544 + 2508 = 3811
Using the maximum \( x \):
x = 33
y = 16
z = 66
6. Final Answer:
- 33 servings of size 1,
- 16 servings of size 2,
- 66 servings of size 3.
x = 33
y = 16
z = 66
2. Express constraints in terms of inequalities:
3. Substitute the minimum values for y and z based on x:
4. Replace y and z in the weight constraint:
5. Calculate the maximum total profit:
- For \( x = 5 \):
Total Profit:
- For \( x = 33 \):
Total Profit:
Using the maximum \( x \):
6. Final Answer:
- 33 servings of size 1,
- 16 servings of size 2,
- 66 servings of size 3.
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