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Question 904149: A farming cooperative mixes two brands of cattle feed. Brand X costs $25 per bag and contains 2 units of nutritional element A, 2 units of element B, and 2 units of element C. Brand Y costs $20 per bag and contains 1 unit of nutritional element A, 9 units of element B, and 3 units of element C. The minimum requirements for nutrients A,B,& C are 12 units, 36 units, and 24 units, respectively. Find the number of bags of each brand that should be mixed to produce a mixture having a minimum cost per bag. What is the minimum cost per bag?
Please show all your work. I am really bad with Linear Programming.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let x equal the number of bags of Brand X.
let y equal the number of bags of Brand Y.
The total cost will be 25*x + 20*y
That's your objective function that you want to minimize.
your constraints are given in the following table:
Nutritional Element Brand X Brand Y
A 2 1
B 2 9
C 2 3
A >= 12 units
B >= 36 units
C >= 24 units
Your constraint equations are:
2x + y >= 12 units of nutritional element A
2x + 9y >= 36 units of nutritional element B
2x + 3y >= 24 units of nutritional element C
There are two other constraints that are implied and also need to be considered.
They are:
x >= 0
y >= 0
This is because the number of bags can never be less than 0.
You need to graph your constraints and then find the feasible region that satisfies all the constraints and then find the corners of the feasible region.
Your minimum cost will be at these corners.
You will graph the equality portion of the constraints.
Therefore you will graph:
2x + y = 12 units of nutritional element A
2x + 9y = 36 units of nutritional element B
2x + 3y = 24 units of nutritional element C
x = 0
y = 0
Once you do that, you will find the area of the graph that satisfies all the inequality constraints.
We'll do the graph first and then find the feasible region.
The graph will look like this:
Now you look for the feasible region.
The feasible region is the area of the graph that satisfies all the inequality equations.
The feasible area will be the area on the graph that is:
on or above the line y = 12 - 2x
on or above the line y = (36-2x)/9
on or above the line y = (24-2x)/3
on or above the line y = 0
on or to the right of the line x = 0
you would have to look at all the lines and determine in which direction the area of feasibility lies.
I could do it manually but it will take too much time and a lot of effort to graph it manually and why do that when I have software to do it for me.
I will show you where the area is so you can see it.
You should look to determine what the area is manually from the first graph I showed you.
The feasible region is the area on the graph that is the darkest shade of gray.
You can see that it is the area that is above all of the constraint lines on the graph.
The maximum / minimum solution will be at the corners of that feasible region which are shown on the graph with their coordinate points.
What's left is to find the minimum cost solution.
That will be at the coordinate points of:
x y 25x + 20y
0 12 25*0 + 20*12 = 0 + 240 = 240
3 6 25*3 + 20*6 = 75 + 120 = 195 *****
9 2 25*9 + 20*2 = 225 + 40 = 265
18 0 25*18 + 20*0 = 450 + 0 = 450
The minimum cost is when he makes 3 bags of brand X and 6 bags of brand Y.
All the constraints have to be met as well.
the constraint equations are:
2x + y >= 12 units of nutritional element A
2x + 9y >= 36 units of nutritional element B
2x + 3y >= 24 units of nutritional element C
x >= 0
y >= 0
When x = 3 and y = 6:
2x + y becomes 2*3 + 1*6 = 6 + 6 = 12 so the first constraint is met.
2x + 9y becomes 2*3 + 9*6 = 6 + 54 = 60 so the second constraint is met.
2x + 3y becomes 2*3 + 3*6 = 6 + 18 = 24 so the third constraint is met.
x and y are both greater than 0 so the fourth and fifth constraints are also met.
Looks like the solution is valid.
The minimum cost is when he produces 3 bags of Brand X and 6 bags of brand Y at a total cost of $195.00.
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