9514 1404 393
Answer:
- no
- 0.2a+0.4b≤2.00; 7a+3b≤30; 6a+9b≥50
Step-by-step explanation:
1. A lunch with 4A and 3B will cost ...
4($0.20) +3($0.40) = $0.80 +1.20 = $2.00
The amount of sugar in that lunch is ...
4(7 g) +3(3 g) = (28 +9) g = 37 g
The amount of protein in that lunch is ...
4(6 g) +3(9 g) = (24 +27) g = 51 g
Comparing these amounts to the goals for the lunch, we see that ...
- the cost constraint of $2.00 maximum is met
- the sugar constraint of 30 g maximum is not met
- the protein constraint of 50 g minimum is met
A lunch consisting of 4 servings of A and 3 servings of B cannot meet all of the constraints.
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2. Evaluating the contributions to cost, sugar, and protein in the first part has shown us how to do the second part.
0.20a +0.40b ≤ 2.00 . . . . . . . cost constraint
7a + 3b ≤ 30 . . . . . . . . . . . . . . .sugar constraint
6a + 9b ≥ 50 . . . . . . . . . . . . . . protein constraint
These are the inequalities that are wanted.
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<em>Additional comment</em>
The attached graph is a graph of these constraints. Solutions to all three inequalities would lie in an area where the three solution spaces overlap. This graph shows there is <em>no such area</em>. All three constraints cannot be met by any choice of foods A and B.