SOLUTION: A dealer has 7600 pounds of peanuts, 5800 pounds of almonds, and 3000 pounds of cashews to be used to make two mixtures. The first mixture wholesales for $8.44 per pound and consis

Question 1110667: A dealer has 7600 pounds of peanuts, 5800 pounds of almonds, and 3000 pounds of cashews to be used to make two mixtures. The first mixture wholesales for $8.44 per pound and consists of 60% peanuts, 30% almonds, and 10% cashews. The second mixture wholesales for $3.17 per pound and consists of 20% peanuts, 50% almonds, and 30% cashews. How many pounds of each mixture should be made to maximize revenue? Find the maximum revenue.

Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52805)   (Show Source): You can put this solution on YOUR website!
.

Let X and Y be the amounts (in pounds) of each mixture. The Revenue function is Z = 8.44*X + 3.17*Y. The constraints are these inequalities: 0.6X + 0.2Y <= 7600 (1) (peanuts) 0.3X + 0.5Y <= 5800 (2) (almonds) 0.1X + 0.3Y <= 3000 (3) (cashews) You need to find the maximum of the objective function under these restrictions (1), (2), (3) and X >= 0, Y>= 0. At this point, the formulation/(the setup) of the linear optimization problem is COMPLETED. Further, you can apply the Linear programming method and solve the problem using a standard Geometry visualization approach.

On how to do it, you can learn from the lesson
    - Solving minimax problems by the Linear Programming method
in this site.

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

Let x be the number of pounds of the first mixture and y be the number of pounds of the second. Then
(1) he has 7600 pounds of peanuts
(2) ... and 5800 pounds of almonds
(3) ... and 3000 pounds of cashews

He wants to maximize his revenue, given that the first mixture sells at $8.44 per pound and the second sells at $3.17 per pound. So the objective function to be maximized is

In slope-intercept form, the three constraint equations are
(1)
(2)
(3)

Algebra or a graphing calculator show that the vertices of the feasibility region are
(0,0)
(38000/3,0)
(11000,5000)
(6000,8000)
(0,10000)

Evaluating the objective function at each of those vertices shows the maximum revenue is $108,690 at (11000,5000).
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