Algebra Question

Question

A store has two types of animal feed available. Type A contains  pounds of oats and 1pound of corn per bag. Type B contains  pounds of oats and 3pounds of corn per bag. A farmer wants to combine the two types so that the resulting mixture has at least 43pounds of oats and at least 17 pounds of corn. The store only has 14 bags of type A feed and 11 bags of type B feed in stock. Type A costs $3 per bag, and type B costs $4

 per bag. How many bags of each type should the farmer buy to minimize her cost?

Answer 1

Objective Function: Cost = 3A + 4B. We want to minimize Cost

Oats Constraint:

43 <= 2A + 2B

A >= -B + 43/2 Write A = f(B). Like y = mx + b

Corn Constraint:

17 <= A + 3B.

A >= -3B + 17. Write A = f(B) Like: y = mx + b

Also; Limited Supplies

A <= 14

B <= 11

Graph all the constraints on one graph, where B is the independent axis and A is the dependent axis.

In other words, let A be like "y" and let B be like "x"

After graphing the four inequalities, you will find a triangular feasible region.

The three corners of the feasible region are:

(B,A) = (11,15), (11, 10.5), (6.5, 15)

NOTE: To determine the coordinates of the above, three points,

you will need to determine the intersection of

the first constraint: A >= -B + 43/2 and

the two constraints for limited supplies.

It turns out that the Corn constraint is irrelevant.

The Corn constraint does not affect the feasible region.

After you find the feasible region,

and the corners of that feasible region,

EVALUATE the Objective function at those points.

(B,A) = (11,15) ==> Cost = 3(15) + 4(11) = 89

(B,A) = (11,10,5) ==> Cost = 3(10.5) + 4(11) = 75.5

(B,A) = (6.5, 15) ==> Cost = 3(15) + 4(6.5) = 71

The cost is minimized at (B,A) = (6.5, 15)

At that point, the cost is $71

That's 6.5 bags of Type B and 15 bags of Type A

If you can't buy half-bags, then adjust the coordinates,

making sure the integers are within the feasible region.

If integer values are required, evaluate the following three points:

(B,A) = (11,15) ==> Cost = 3(15) + 4(11) = 89

(B,A) = (11,11) ==> Cost = 3(11) + 4(11) = 77

(B,A) = (7, 15) ==> Cost = 3(15) + 4(7) = 73

If you can NOT buy half-bags, (must use integer values)

The cost is minimized at (B,A) = (7, 15)

At that point, the cost is $73

That's 7 bags of Type B and 15 bags of Type A