Question

Joey is buying plants for his gardenHe wants to have at least twice as many flowering plants as nonflowering plants and a minimum of 36 plants in his gardenFlowering plants sell for $8, and nonflowering plants sell for $5Joey wants to purchase a combination of plants that minimizes costLet x represent the number of flowering plants and y represent the number of nonflowering plants.

What are the vertices of the feasible region for this problem?

Answer 1

Answer:

The vertices of Feasible region are $(24,12) and (0,36)$.

Step-by-step explanation:

Given that,

He wants to have at least twice as many flowering plants as nonflowering plants.

Let  number of flowering plants be $x$ .

and number of non-flowering plant be $y$.  

According to question,

Inequality form by from given situation is :

                                       $x\geq 2y$                       ..............(1)

 And Minimum number of flowers should in Garden must be represented by the Inequality is :

                                 $x+y\geq 36$                      .................(2)

Cost of x flowering plants be  $8x$.

Cost of y nonflowering plants be $9y$

Also, he has to minimize the cost then Inequality will be  

                Minimum cost  $C= 8x+5y$

Now,

The feasible solution are one who satisfy the constraints (1) & (2).

So, Solving Constraints (1) & (2) we get

putting the value of equation (1) in equation (2)

                            $2y + y = 36$

                                  $3y=36$

                                   $y=12$

then                            $x=24$

Again Solving Equation (1) & (1) we get, $x=0$ & $y=36$

Hence,

The vertices of Feasible region are $(24,12) and (0,36)$.