Find out how to make the most profit..

Question

Need some help getting started with this one please..

The facility can make at most 100 tires a day. Each large tire requires 2 lbs of rubber, and each small tire requires 1 lb of rubber. There is 500 lbs of rubber available. The income from each large tire is $50 and from each small tire is $20. How many of each size tire should be made to maximize income?

Answer 1

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.

 

 

STEP 1: Define variables

 

Here, your variables are number of large tires and number of small tires:

 

x = # large tires

y = # small tires

 

 

STEP 2: Write inequalities that bound production

 

x ≥ 0 (you can't make a negative number of large tires)

y ≥ 0 (you can't make a negative number of small tires)

 

x + y ≤ 100 (the facility can't make more than 100 total tires per day)

 

2x +1y ≤ 500 (the facility can't use more than 500 lbs. of rubber; large tires take 2 lbs. while small tires take 1 lb.)

 

 

STEP 3: Graph each inequality and find the region of overlap (solution set)

 

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 y-axis)

y = 0 (the x-axis)

y = -x + 100 (a diagonal line with y-intercept 100 and slope -1)

y = -2x + 500 (a diagonal line with y-intercept 500 and slope -2)

 

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 that 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 = -x + 100 (since that's where x + y ≤ 100), and everything below y = -2x + 500 (since that's where 2x + y ≤ 500).

 

The solution set for this problem is a triangle bounded by the two axes and y = -x + 100. This is unusual, since it means the equation we wrote for the pounds of rubber was totally useless, but...oh well.

 

 

STEP 4: Find points of intersection

 

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.

 

Luckily, since you don't have to find the intersection of y = -x + 100 and y = -2x + 500 (that's not one of the vertices of the solution set), all of your solution set's vertices are pretty straightforward.

 

x = 0 and y = 0:

x and y are already given. The lines intersect at (0, 0).

 

x = 0 and y = -x + 100:

Plug 0 into the second equation for x to get y = -0 + 100 = 100. The lines intersect at (0, 100).

 

y = 0 and y = -x + 100:

Plug 0 into the second equation for y to get 0 = -x + 100, so -x = -100 and x = 100. The lines intersect at (100, 0).

 

And, in case you're curious, y = -x + 100 and y = -2x + 500:

Plug (-x + 100) into the second equation for y to get -x + 100 = -2x + 500, so x + 100 = 500 and x = 400. Then plug x = 400 into the first equation to get y = -400 + 100 = -300. The lines intersect at (400, -300). Again, though, we don't need this point. It's not one of the vertices of the triangular solution set.

 

 

STEP 5: Write an equation for profit and test each vertex for maximization

 

Now, we'll finally take into account the income from each tire type. Since large tires make $50 each and small tires make $20 each, we can represent profit P as:

 

P = 50x + 20y

 

Our vertices are (0, 0), (100, 0), and (0, 100). One of these will maximize the profit equation. Plug the values in for x and y to find out which:

 

(0, 0) yields P = 50(0) + 20(0) = 0 + 0 = $0

(100, 0) yields P = 50(100) + 20(0) = 5000 + 0 = $5,000

(0, 100) yields P = 50(0) + 20(100) = 0 + 2000 = $2,000

 

Since (100, 0) maximizes profit, the facility should spend all its time making large tires. It should make 100 large tires and 0 small tires each day.

 

That's such a straightforward answer because the amount of rubber the facility has doesn't actually matter. They do have enough rubber to make 100 large tires every day, so they might as well do so. You may want to check to make sure you have the correct numbers, because usually these problems don't give you useless bounds.

 

 

You can follow these steps for any linear programming problem you're given. Hope this helps!