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A manufacturer makes two items, A and B. Item A requires 3 minutes of labor to assemble and B requires 4 minutes of assembly time.
Item A costs $2 in raw materials and B costs $1. There is a maximum of 3,000 labor minutes available for assembly and a budget
of $1,000 in raw material costs per day. Assuming they sell all the produce and that the profit is $5 per item A and $4 per item B,
how many of each item must be produced in order to maximize profit?
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Let X = # items A; Y = # items B.
From the condition, we have this formulation of maximization problem:
(1) the objective function to maximize is the profit P = 5X + 4Y dollars.
Restrictions
(2) 3X + 4Y <= 3000 minutes (assembly time)
(3) 2X + Y <= 1000 dollars (material cost)
(4) X >= 0, Y >= 0.
You can make a plot of the feasibility domain.
It is a quadrilateral in QI with the vertices (X,Y) = (0,0), (500,0), (200,600), (0,750).
The solution is one of these 4 points, where the objective function (profit) has a maximum.
You calculate the values of the function P(X,Y) at listed points
P(0,0) = 0
P(500,0) = 5*500 + 4*0 = 2500
P(200,600) = 5*200 + 4*600 = 3400
P(0,750) = 5*0 + 4*750 = 3000.
Comparing these values, you find the optimal point.
It is (X,Y) = (200,600), 200 items A and 600 items B, providing maximum profit of 3400 dollars.
Solved.
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In this site, there is a lesson
- Solving minimax problems by the Linear Programming method
which explains, for beginners, metodology of solving such problems in more details.