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)

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 and the number of Spring bouquet is .

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

and the total number of daisies required

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

(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:  

If we graph the constraints now, then the vertices of the common shaded region are:       and

For (0, 0) ⇒  

For (12, 0) ⇒  

For (9, 6) ⇒  

For (0, 15) ⇒  

So, the profit will be maximum when   and

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

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

add -60 to 120 which will result in : 2x+3y=180

then subtract 2x: 3y=-2x+180

lastly, divide by 2: y= -2/3x+60

hope this is the answer you're looking for: )