Answer:
The florist should make 9 Easter bouquet and 6 Spring bouquet to maximize the profit.
Step-by-step explanation:
Suppose, the number of the Easter bouquet is [tex]x[/tex] and the number of Spring bouquet is [tex]y[/tex].
The Easter bouquet requires 10 jonquils and 20 daisies, and the Spring bouquet requires 5 jonquils and 20 daisies.
So, the total number of jonquils required [tex]=10x+5y[/tex]
and the total number of daisies required [tex]=20x+20y[/tex]
Given that, there are total 120 jonquils and 300 daisies are available. So, the constraints will be........
[tex]10x+5y\leq 120;\\ \\ 20x+20y\leq 300;\\ \\ x\geq 0; y\geq 0[/tex]
(As the number of each type of bouquet can't be negative)
Now, each Easter bouquet produces a profit of $1.50 and each Spring bouquet produces a profit of $1. So, the profit function will be: [tex]P=1.50x+1y[/tex]
If we graph the constraints now, then the vertices of the common shaded region are: [tex](0,0), (12,0), (9,6)[/tex] and [tex](0,15)[/tex]
For (0, 0) ⇒ [tex]P=1.50(0)+1(0)=0[/tex]
For (12, 0) ⇒ [tex]P=1.50(12)+1(0)=18[/tex]
For (9, 6) ⇒ [tex]P=1.50(9)+1(6)=13.50+6= 19.50[/tex]
For (0, 15) ⇒ [tex]P=1.50(0)+1(15)=15[/tex]
So, the profit will be maximum when [tex]x=9[/tex] and [tex]y=6[/tex]
Thus, the florist should make 9 Easter bouquet and 6 Spring bouquet to maximize the profit.
