Question 1118954: A manufacturer of golf clubs makes a profit of $40 per set on a model A set and $60 per set on a model B set. Daily production of the model A clubs is between 30 and 60 sets, inclusive, and that of the model B clubs is between 10 and 30 sets, inclusive. The total daily production is not to exceed 60 sets. How many sets of each model should be manufactured per day to maximize the profit?
Answer by ikleyn(52786) (Show Source):
You can put this solution on YOUR website! .
Let X and Y be the numbers of model A sets and model B sets, respectively.
Then the constrains are
30 <= X <= 60,
10 <= Y <= 30,
X + Y <= 60.
The profit function is C(X,Y) = 40*X + 60*Y dollars to maximize.
The feasibility area is shown in the Figure below
Plots X = 30 (red), X = 60 (green), Y = 10 (blue), Y = 30 (magenta) and X + Y = 60.
The feasibility domain is the triangle area bounded by the red, blue and sloped lines.
The points to check the cost function are
P1 = (30,10) (right angle vetex)
P2 = (30,30) (upper vertex of the triangle)
P3 = (50,10) (most right vertex of the triangle).
The values of the profit function are
at P1: P(30,10) = 30*40 + 10*60 = 1800;
at P2: P(30,30) = 30*40 + 30*60 = 3000;
at P3: P(50,10) = 50*40 + 10*60 = 2600.
You are looking for the maximum - hence, your solution is at P2.
Answer. 30 model A sets and 30 model B sets is the solution which provides the maximal profit of $3000 per day.
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To see other similar problems solved by the same method, look into the lesson
- Solving minimax problems by the Linear Programming method
in this site.
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