Answer:
A. Ms. Waller should make 15 posy bouquets and 0 cascade bouquets to maximize her profit.
Step-by-step explanation:
Let x be the number of posy bouquet and y be the number of cascade bouquet.
Given,
The time taken for making a posy bouquet = 15 minutes,
And, the time taken for making a cascade bouquet = 20 minutes,
Thus, the total time taken = 15x + 20y,
Since, no more than 300 minutes is taken to make the bouquets,
⇒ 15x + 20y ≤ 300
Now, each posy bouquet must have 12 stems and each cascade bouquet must have 6 stems,
So, the total number of stems = 12x + 6y,
There must be no more than 180 stems in all,
⇒ 12x + 6y ≤ 180,
Now, we have to maximize the profit if each posy bouquet has $ 10 profit and each cascade bouquet has $ 7 profit.
⇒ Total profit = 10x + 7y,
Now, the numbers of bouquets can not be negative,
⇒ x ≥ 0 and y ≥ 0
Thus, there is a LPP that represents the given situation for which the function that must be maximize is,
Z = 10x + 7y,
And, the subject of constraints are,
15x + 20y ≤ 300; 12x + 6y ≤ 180,
x ≥ 0, y ≥ 0
By graphing the above inequalities,
We found that,
The vertex points of the feasible region are,
(0,0), (15,0) and (0,20),
Since, at (0,0), the profit would be,
Z = 10 × 0 + 7 × 0 = 0 + 0 = 0
At (15,0), the profit would be,
Z = 10 × 15 + 7 × 0 + 150 + 0 = 150
At (0,20), the profit would be,
Z = 10 × 0 + 7 × 20 = 0 + 140 = 140
Hence, the profit would be maximum when x = 15 and y = 0,
That is, Ms. Waller should make 15 posy bouquets and 0 cascade bouquets to maximize her profit.
Option A is correct.
