Question 1071240: 2 A small manufacturer produces two kinds of good, A and B, for which demand exceeds capacity. The production costs for A and B are $6 and $3, respectively, each, and the corresponding selling prices are $7 and $4. In addition, the transport costs are 20 cents and 30 cents for each good of type A and B, respectively. The conditions of a bank loan limit the manufacturer to maximum weekly production costs of $2700 and maximum weekly transport costs of $120.
How should the manufacturer arrange production to maximize profit?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! x = number of units of product A.
y = number of units of product B.
your constraint inequalitiess are:
x >= 0
y >= 0
6x + 3y <= 2700
.2x + .3y <= 120
selling price is equal to 7x + 4y
profit is equal to 7x - 6x - .2x + 4y - 3y - .3y
combine like terms and you get:
profit is equal to .8x + .7y
that's your objective function.
using the desmos.com calculator, you would graph the opposite inequalities.
to be more specific, you would graph the following inequalities.
x <= 0
y <= 0
6x + 3y >= 2700
.2x + .3y >= 120
the area of the graph that is NOT shaded is your region of feasibility.
your graph will look like this.
you then find the corner points of this region and evaluate your objective function at those corner points.
you will get.
profit at (0,400) = 280
profit at (375,150) = 405
profit at (450,0) = 360
your maximum profit is at (375,150).
you have to meet your constraints at those corner points.
production costs at (375,150) = 6x + 4y = 2700 which meets the constraint that they be less than or equal to 2700.
transport costs at (375,150) = .2x + .3y = 120 which meets the constraint that they be less than or equal to 120.
x >= 0 and y >= 0 are also meet the constraint that they be greater than or equal to 0 at the point (375,150).
all constraints are met.
your maximum profit is when you sell 375 units of product A and 150 units of product B.
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