36 jars of salsa
Step-by-step explanation:
There are many ways you can solve this. I like to look to see if there is an obvious solution. Here, if we were to consider the case of having 10 tomatoes only, we would need 1 onion for sauce, and we would make $2. Using the same 10 tomatoes in salsa, we would need 1/2 onion, and we would make $3. It seems that the profit per onion is considerably greater for salsa than for sauce. So, to maximize profit, we should restrict our production to salsa only.
Considering the inventory of tomatoes and onions, we find that we have enough tomatoes for 36 jars of salsa, and enough onions for 60 jars of salsa. So, the limit on our profit from salsa is the number of tomatoes we have. The 36 jars we can make will easily satisfy our requirement for at least 20 jars.
To maximize profit, we should make ...
0 jars of tomato sauce36 jars of salsa
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Another way to solve this is to graph the solution space and identify the combination of sauce and salsa that maximizes profit.
Let x = jars of tomato sauce; y = jars of salsa. The conditions imposed by the problem statement are ...
x + y ≥ 20 . . . . . . we must make at least 20 total jars
10x +5y ≤ 180 . . . the number of tomatoes we use cannot exceed 180
1x + 1/4y ≤ 15 . . . . the number of onions we use cannot exceed 15
The graph of these inequalities shows vertices of the feasible region near ...
(13, 7), (12, 12), (0, 36)
The profit for each of these mixes are, respectively,
2·13 +1.5·7 = 36.502·12 +1.5·12 = 42.002·0 +1.5·36 = 54.00 (maximum profit)
The making of 0 jars of tomato sauce and 36 jars of salsa will maximize profit.
