How many units of each product should be produced in each shift to maximize the company's profit? (x
Mathematics

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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 $4/unit of Product A and a profit of $7/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? (x, y) =
What is the optimal profit? (Round your answer to the nearest whole number.) $
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This is a type of problem called "linear programming," where you typically plot a system of inequalities (each of which represents a bound on the production of goods) and then test specific values to figure out how to maximize or minimize a company's profits or costs.
STEP 1: Define variables
Here, your variables are number of units of Product A and number of units of Product B:
x = # Product A
y = # Product B
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This is a type of problem called "linear programming," where you typically plot a system of inequalities (each of which represents a bound on the production of goods) and then test specific values to figure out how to maximize or minimize a company's profits or costs.
STEP 1: Define variables
Here, your variables are number of units of Product A and number of units of Product B:
x = # Product A
y = # Product B
STEP 2: Write inequalities that bound production for Machine I and Machine II
x ≥ 0 (you can't make a negative number of units of Product A)
y ≥ 0 (you can't make a negative number of units of Product B)
6x + 9y ≤ 600 (the company can use Machine I for 10 hours = 600 minutes each shift; Product A takes 6 minutes on Machine I while Product B takes 9 minutes)
x ≥ 0 (you can't make a negative number of units of Product A)
y ≥ 0 (you can't make a negative number of units of Product B)
5x + 4y ≤ 360 (the company can use Machine II for 6 hours = 360 minutes each shift; Product A takes 5 minutes on Machine II while Product B takes 4 minutes)
STEP 3: Graph each inequality and find the region of overlap (solution set) forMachine I and Machine II
First, change each inequality so it's in slope intercept form. Then, ignore the inequality signs for the moment and just graph the system of inequalities as though they were normal equations:
x = 0 (the yaxis)
y = 0 (the xaxis)
y = (2/3)x + 200/3 (a diagonal line with yintercept 200/3 and slope 2/3)
x = 0 (the yaxis)
y = 0 (the xaxis)
y = (5/4)x + 90 (a diagonal line with yintercept 90 and slope 5/4)
Finally, figure out which region on the graph represents the solution set of the inequalities. You can do this by shading the half of the graph "above" or "below" each inequality based on which half contains solutions to the given inequality. Then, whichever region has been shaded for every single inequality is your solution set. Usually, it's an enclosed polygon shape.
Here, you would shade everything to the right of x = 0 (since that's where x ≥ 0), everything above y = 0 (since that's where y ≥ 0), everything below y = (2/3)x + 200/3 (since that's where 6x + 9y ≤ 600), and everything below y = (5/4)x + 90(since that's where 5x + 4y ≤ 360).
The solution set for Machine I is a triangle bounded by the two axes and y = (2/3)x + 200/3.
The solution set for Machine II is a triangle bounded by the two axes and y = (5/4)x + 90.
STEP 4: Find points of intersection for Machine I and Machine II
Minimizing or maximizing the profit or costs requires you to find the vertices of that solution set's polygon. The maximum or minimum value will always be one of those vertices. To find the vertices, you'll need to know where each pair of lines intersects. Some vertices are straightforward to find, some are not as straightforward.
x = 0 and y = 0:
x and y are already given. The lines intersect at (0, 0).
x = 0 and y = (2/3)x + 200/3:
Plug 0 into the second equation for x to get y = (2/3)0 + 200/3 = 0 + 200/3 = 200/3. The lines intersect at (0, 200/3).
y = 0 and y = (2/3)x + 200/3:
Plug 0 into the second equation for y to get 0 = (2/3)x + 200/3, or 2/3x = 200/3. This means that 2x = 200, or x = 100. The lines intersect at (100, 0).
x = 0 and y = (5/4)x + 90:
Plug 0 into the second equation for x to get y = (5/4)0 + 90 = 0 + 90 = 90. The lines intersect at (0, 90).
y = 0 and y = (5/4)x + 90:
Plug 0 into the second equation for y to get 0 = (5/4)x + 90, so (5/4)x = 90 and 5x = 360. That means that x = 72, so the lines intersect at (72, 0).
STEP 5: Write an equation for profit and test each vertex for maximization ofMachine I and Machine II
Now, we'll finally take into account the income from each product. Since Product A's make $3 each and Product B's make $4 each, we can represent profit P as:
P = 3x + 4y
Our vertices for Machine I are (0, 0), (0, 200/3), and (100, 0). Our vertices for Machine II are (0, 0), (0, 90), and (72, 0). One of these will maximize the profit equation for each machine. Plug the values in for x and y to find out which:
(0, 0) yields P = 3(0) + 4(0) = 0 + 0 = $0
(0, 200/3) yields P = 3(0) + 4(200/3) = 0 + 800/3 = $266.67
(100, 0) yields P = 3(100) + 4(0) = 300 + 0 = $300
(0, 0) yields P = 3(0) + 4(0) = 0 + 0 = $0
(0, 90) yields P = 3(0) + 4(90) = 0 + 360 = $360
(72, 0) yields P = 3(72) + 4(0) = 216 + 0 = $216
The maximum profit for Machine I is $300 for (100, 0). The maximum profit for Machine II is $360 for (0, 90).
Since 100 Product A's maximize profit for Machine I and 90 Product B's maximize profit for Machine II, the company should make 100 units of Product A and 90 units of Product B each shift. The profit for that pattern of manufacturing is $300 + $360 = $660 each shift.
Machine I will spend its 600 minutes making 100 units of Product A. This will take 6(100) = 600 minutes. So, there are 0 minutes remaining on Machine I. Machine II will spend its 360 minutes making 90 units of Product B. This will take 4(90) = 360 minutes. So, there are 0 minutes remaining on Machine II.
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