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Question 1137186: A farmer wants to see that his herd gets the minimum daily requirement of three basic nutrients A, B and C. Daily requirements are 14 for A, 12 for B and 18 for C. Product y1 has two units of A, and one unit each of B and C; product y2 has one unit each of A and B, and three units of C. The cost of y1 is $2 and the cost of y2 is $4.Determine the least-cost combination of y1 and y2 .
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let x = the number of y1 products.
let y = the number of y2 products.
product y1 has 2 units of A and 1 unit of B and 1 unit of C.
product y2 has 1 unit of A and 1 unit of B and 3 units of C.
the total total requirement is for 14 units of A and 12 units of B and 18 units of C.
for the A requirement, your constraint equation is 2x + y >= 14
for the B requirement, your constraint equation is x + y >= 12
for the C requirement, your constraint equation is x + 3y >= 18
another requirement is that x can't be less than 0 and y can't be less than 0.
the inequality for that is x >= 0 and y >= 0.
your objective function is cost = 2 * x + 4 * y.
that's the function you want to minimize.
using the desmos.com calculator, you would graph the opposite of the inequalities.
the area of the graph that is not shaded will be your region of feasibility.
you will then evaluate your objective function at the corner points of this feasible region.
the corner point with the least cost and that satisfies the constraints is your solution.
the inequalities are:
2x + y >= 14
x + y >= 12
x + 3y >= 18
x >= 0
y >= 0
you will graph:
2x + y <= 14
x + y <= 12
x + 3y <= 18
x <= 0
y <= 0
the area of the graph that is not shaded is your region of feasibility.
here's that the graph looks like.
the corner point of the feasible region and the evaluation of the objective function at those corner points is:
the corner points are (0,14), (2,10), (9,3), (18,0).
the objective function is evaluated at each corner point and the corner point with the least cost that satisfies the constraints is the solution.
that point is (9,3) where the total cost is 2 * 9 + 4 * 3 = 18 + 12 = 30.
you can check for yourself to see that this corner point has the least cost.
the constraints have to all be satisfied as well.
at the corner point of (9,3), .....
the number of units of A are 2 * 9 + 1 * 3 = 21 which is greater than or equal to 14.
the number of units of B are 1 * 9 + 1 * 3 = 12 which is greater than or equal to 12.
the number of units of C are 9 * 1 + 3 * 4 = 21 which is greater than or equal to 18.
the cost is minimum and all the constraints are satisfied, so you would buy 9 product y1 and 3 product y2 and your requirements will be satisfied and your cost will be minimum.
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