Question

flowers unlimited has two spring floral arrangements, the easter bouquet and the spring bouquet. the easter bouquet requires 10 jonquils and 20 daisies and produces a profit of $1.50. the spring bouquet required 5 jonquils and 20 daisies and yield a profit of $1. how many of each type of arrangement should the florist make to maximize the profit if 120 jonquils and 300 daisies are available? (assume that all bouquets will be sold)

Answer 1

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 $x$ and the number of Spring bouquet is $y$.

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 $=10x+5y$

and the total number of daisies required $=20x+20y$

Given that, there are total 120 jonquils and 300 daisies are available. So, the constraints will be........

$10x+5y\leq 120;\\ \\ 20x+20y\leq 300;\\ \\ x\geq 0; y\geq 0$

(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:  $P=1.50x+1y$

If we graph the constraints now, then the vertices of the common shaded region are:      $(0,0), (12,0), (9,6)$ and $(0,15)$

For (0, 0) ⇒  $P=1.50(0)+1(0)=0$

For (12, 0) ⇒  $P=1.50(12)+1(0)=18$

For (9, 6) ⇒  $P=1.50(9)+1(6)=13.50+6= 19.50$

For (0, 15) ⇒  $P=1.50(0)+1(15)=15$

So, the profit will be maximum when  $x=9$ and $y=6$

Thus, the florist should make 9 Easter bouquet and 6 Spring bouquet to maximize the profit.