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.
__
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.
_____
Additional comment
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 no such area. All three constraints cannot be met by any choice of foods A and B.
