SOLUTION: I have a graph that I am supposed to fill out along with answering the equation. The problem is this: I need to have some stuffed animals for a kid’s party. I want to have 24

Algebra ->  Linear-equations -> SOLUTION: I have a graph that I am supposed to fill out along with answering the equation. The problem is this: I need to have some stuffed animals for a kid’s party. I want to have 24       Log On


   

Question 727599: I have a graph that I am supposed to fill out along with answering the equation. The problem is this:
I need to have some stuffed animals for a kid’s party. I want to have 24 cats, 20 dogs and 12 dragons. (If I end up with extras, that is fine.) To save money, I will buy the big party assortments. There are two assortments available. The Stuffed-Animals-A-Plenty, for $5 has 3 cats, 2 dogs, and 4 dragons. The Kids-Box, for $6 has 4 cats, 4 dogs, and 3 dragons. How many of each party assortment should I buy to minimize my cost?
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the number of Stuffed-Animals-A-Plenty's, which means we have
3x cats, 2x dogs, and 4x dragons, at a cost of $5x

Let y = the number of Kids-Box's, which means we have
4y cats, 4y dogs, and 3y dragons, at a cost of $6y

3x + 4y ≧ 24     <---- the cat inequality
2x + 4y ≧ 20     <---- the dog equality
4x + 3y ≧ 12     <---- the dragon equality
      x ≧ 0      <---- x cannot be negative
      y ≧ 0      <---- y cannot be negative

The objective function (money equation) to minimize is

C(x,y) = 5x + 6y   

The equations of the lines are

3x + 4y = 24     <---- the boundary equation for the cats
2x + 4y = 20     <---- the boundary equation for the dogs
4x + 3y = 12     <---- the boundary equation for the dragons. 



Using the origin as a test point shows that all solutions
are above all three lines. The blue line can be eliminated
since it is below both of the other lines. 

Solve the system

3x + 4y = 24    
2x + 4y = 20   

and get solution x=4, y=3 which is the point (4,3).  So
the corner points are the y-intercept of the red line,
which is (0,6) and the x-intercept of the green line, which is (10,0)

The feasible region consists of all the points on or above these lines.



We evaluate the objective function at the three corner points:

Corner | 
Point  |  C(x,y) = 5x + 6y
---------------------------------
 (0,6) |  C(0,6) = 5·0  + 6·6 =  0 + 36 = 46
 (4,3) |  C(4,3) = 5·4  + 6·3 = 20 + 18 = 38 
(10,0) | C(10,0) = 5·10 + 6·0 = 50 +  0 = 50

Therefore the minimum cost is $38 when 4 Stuffed-Animals-A-Plenty's
and 3 Kids-Boxes are purchased.

Edwin