SOLUTION: A company manufactures x units of Product A and y units of Product B, on two machines, I and II. It has been determined that the company will realize a profit of $3/unit of Product

Question 718370: A company manufactures x units of Product A and y units of Product B, on two machines, I and II. It has been determined that the company will realize a profit of $3/unit of Product A and a profit of $4/unit of Product B. To manufacture a unit of Product A requires 6 min on Machine I and 5 min on Machine II. To manufacture a unit of Product B requires 9 min on Machine I and 4 min on Machine II. There are 5 hr of machine time available on Machine I and 3 hr of machine time available on Machine II in each work shift. How many units of each product should be produced in each shift to maximize the company's profit? THANKS
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
First of all let's consider the constraints/limits on the values for x and y:

and since it doesn't make sense to produce a negative number of units.
We are limited to 5 hours (or 300 minutes) on Machine I.
- To produce 1 unit of Product A it takes 6 minutes on Machine I. To produce x units of Product A it will take 6x minutes on machine I.
- To produce 1 unit of Product B it takes 5 minutes on Machine I. To produce y units of Product B it will take 5y minutes on machine I.
- Together we will need 6x + 5y minutes on Machine I to produce x units of Product A and y units of Product B.
So
We are limited to 3 hours (or 180 minutes) on Machine II.
- To produce 1 unit of Product A it takes 5 minutes on Machine II. To produce x units of Product A it will take 5x minutes on machine II.
- To produce 1 unit of Product B it takes 4 minutes on Machine II. To produce y units of Product B it will take 4y minutes on machine II.
- Together we will need 5x + 4y minutes on Machine II to produce x units of Product A and y units of Product B.
So

So altogether the constraints on x and y are:
and and and
To draw the graph of this, so we can see the x's and y's that are allowed, graph the four related equations:

x = 0 (which is the y-axis)
y = 0 (which is the x-axis)
6x + 5y = 300
5x + 4y = 180

There will be 4 points of intersection between these lines:

The origin (where x=0 intersects with y=0)
The x intercept of 5x + 4y = 180 (where y=0 intersects 5x + 4y = 180)
The y intercept of 6x + 5y = 300 (where x=0 intersects 6x + 5y = 300)
The point where 5x + 4y = 180 and 6x + 5y = 300 intersect.

A quadrilateral (four-sided polygon) is formed with these points of intersection as vertices. Any point in this quadrilateral, including the sides, have coordinates which are acceptable values of x and y in this problem.

More importantly, the maximum profit will occur at one of the 4 vertices of the quadrilateral. (So will the minimum profit, for that matter.) So to find the maximum profit we will need to figure out the coordinates of the 4 vertices. (The origin is obviously (0, 0). The others you will have to figure out.)

The profit from the sale of 1 unit is $3 for Product A and $4 for Product B. The profit from selling x units of Product A will be 3x dollars and the profit from selling y units will be 4y dollars. The total profit, then, will be:
P = 3x + 4y
To find the maximum profit...

For each vertex: Substitute its coordinates for x and y into P = 3x + 4y and get its profit. This will give you 4 values for P (the profit).
The x and y that makes P the biggest is the answer to your problem.

Here's what the graph of the four lines looks like (remember x=0 and y=0 are the y and x axes respectively):

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