how many courses of each type should sydney take if she wants to minimize the cost of her education?

Question

the texas technical institute for turtle tutelage offers classes in language, mathematics, and physics. language courses are worth 3 credit , math courses worth 4 credit , and physics are worth 5 credit. the prise to take a language course is $15, math course is $12, physics course is $18. to graduate, sydney the turtle must take at least 35 classes and accumulate a minimum of 130 credits. in addition, the college requires that a minimum of 5 each type of course be taken. how many courses of each type should sydney take if she wants to minimize the cost of her education?

Answer 1

Most linear programming problems work the same way:

 

 

 

STEP 1: ASSIGN VARIABLES

 

Look at the actual question to help you assign variables. It asks for "how many courses of each type." That means our variables should be:

 

x = number of language courses

y = number of math courses

z = number physics courses

 

 

 

STEP 2: WRITE BOUNDING EQUATIONS

 

Think about upper and lower bounds for each variable, plus any other details they gave us.

 

 

Number of credits per course / minimum of 130 credits:

 

3x + 4y + 5z ≥ 130

 

 

Price per course:

 

Since we're minimizing cost in this problem, we won't use pricing details for the bounding equations. The pricing details will come in handy later.

 

 

At least 35 classes:

 

x + y + z ≥ 35

 

 

Minimum of 5 of each type of course:

 

x ≥ 5

y ≥ 5

z ≥ 5

 

 

So, we have five bounding equations:

 

3x + 4y + 5z ≥ 130

x + y + z ≥ 35

x ≥ 5
y ≥ 5
z ≥ 5

 

 

 

STEP 3: FIND POINTS OF INTERSECTION

 

With a two-variable linear programming problem, this is where we would graph the bounding inequalities and find the solution region. The points of intersection that served as vertices for the solution region would be our possible solutions.

 

It's much harder to graph or visualize three-variable equations, but we'll still use a similar method. We'll pick three bounding equations at a time to find their point of intersection. If that point of intersection is a solution for the other two bounding equations, it's a possible solution to our problem. This will take a while...

 

 

Start with 3x + 4y + 5z = 130, x + y + z = 35, and x = 5.

 

Since x = 5, we have:

 

15 + 4y + 5z = 130 --> 4y + 5z = 115

5 + y + z = 35 --> y + z = 30

 

Solve for y in the second equation to get y = 30 - z, then substitute into the first equation to solve for z:

 

4(30 - z) + 5z = 115

120 - 4z + 5z = 115

z = -5

 

So, y = 30 - (-5) = 35.

 

This point, (5, 35, -5), doesn't solve the inequality z ≥ 5. So we'll ignore it.

 

 

Now try 3x + 4y + 5z = 130, x + y + z = 35, and y = 5.

 

3x + 20 + 5z = 130 --> 3x + 5z = 110

x + 5 + z = 35 --> x + z = 30

 

Therefore, x = 30 - z. So:

 

3(30 - z) + 5z = 110

90 - 3z + 5z = 110

90 + 2z = 110

2z = 20

z = 10

 

Then x = 30 - 10 = 20.

 

The point (20, 5, 10) does fulfill the other two equations, since x ≥ 5 and z ≥ 5. So, this is a possible solution.

 

 

Now try 3x + 4y + 5z = 130, x + y + z = 35, and z = 5.

 

3x + 4y + 25 = 130 --> 3x + 4y = 105

x + y + 5 = 35 --> x + y = 30

 

So, x = 30 - y. Therefore:

 

3(30 - y) + 4y = 105

90 - 3y + 4y = 105

90 + y = 105

y = 15

 

That means that x = 30 - 15 = 15.

 

The point (15, 15, 5) does fulfill the other two inequalities since x ≥ 5 and y ≥ 5, so this is a possible solution.

 

 

Now try 3x + 4y + 5z = 130, x = 5, and y = 5.

 

15 + 20 + 5z = 130

35 + 5z = 130

5z = 95

z = 19

 

The point (5, 5, 19) does fulfill the inequality z ≥ 5. For x + y + z ≥ 35, however, we get 5 + 5 + 19 ≥ 35 --> 29 ≥ 35, which is not true. So, this point is not a possible solution.

 

 

Now try 3x + 4y + 5z = 130, x = 5, and z = 5.

 

15 + 4y + 25 = 130

40 + 4y = 130

4y = 90

y = 22.5

 

The point (5, 22.5, 5) also doesn't fulfill the inequality x + y + z ≥ 35. 5 + 22.5 + 5 ≥ 35 --> 32.5 ≥ 35, which is not true. So, this point is not a possible solution.

 

 

Now try 3x + 4y + 5z = 130, y = 5, and z = 5.

 

3x + 20 + 25 = 130

3x + 45 = 130

3x = 85

x = 28 ¹/₃

 

The point (28 ¹/₃, 5, 5) does fulfill both other inequalities, since x = 28 ¹/₃ ≥ 5 and 28 ¹/₃ + 5 + 5 = 38 ¹/₃ ≥ 35. So, it's a possible solution.

 

 

Now try x + y + z = 35, x = 5, and y = 5.

 

5 + 5 + z = 35

10 + z = 35

z = 25

 

The point (5, 5, 25) does fulfill both other inequalities, since z = 25 ≥ 5 and 3x + 4y + 5z = 15 + 20 + 125 = 160 ≥ 130. So, this is a possible solution.

 

 

Now try x + y + z = 35, x = 5, and z = 5.

 

5 + y + 5 = 35

10 + y = 35

y = 25

 

The point (5, 25, 5) does fulfill both other inequalities, since y = 25 ≥ 5 and 3x + 4y + 5z = 15 + 100 + 25 = 140 ≥ 130. So, this is a possible solution.

 

 

Now try x + y + z = 35, y = 5, and z = 5.

 

x + 5 + 5 = 35

x + 10 = 35

x = 25

 

The point (25, 5, 5) does not fulfill the other inequalities, since 3x + 4y + 5z = 75 + 20 + 25 = 120 is not greater than 130. So, this is not a solution.

 

 

Finally, try x = 5, y = 5, and z = 5.

 

The point (5, 5, 5) clearly does not fulfill the other inequalities, since x + y + z = 5 + 5 + 5 = 15 is not greater than or equal to 35. So, this is not a possible solution.

 

 

That means our possible solutions are:

 

(20, 5, 10)

(15, 15, 5)

(28 ¹/₃, 5, 5)

(5, 5, 25)

(5, 25, 5)

 

 

 

STEP 4: WRITE A COST EQUATION

 

This equation will calculate cost C for a given point (x, y, z) to tell us how much a given combination of classes will cost Sydney. Now, we get to take the course pricing into account:

 

C = 15x + 12y + 18z

 

 

 

STEP 5: TEST POSSIBLE SOLUTIONS TO FIND MINIMUM COST

 

Now, we'll test the possible solutions we got to find out which one minimizes cost. That will be our true solution.

 

C = 15(20) + 12(5) + 18(10) = 300 + 60 + 180 = $540

C = 15(15) + 12(15) + 18(5) = 225 + 180 + 90 = $495

C = 15(28 ¹/₃) + 12(5) + 18(5) = 425 + 60 + 90 = $575

C = 15(5) + 12(5) + 18(25) = 75 + 60 + 450 = $585

C = 15(5) + 12(25) + 18(5) = 75 + 300 + 90 = $465

 

The lowest cost is the point (5, 25, 5). So, Sydney should take 5 language courses, 25 math courses, and 5 physics courses. This makes sense, since math courses are the cheapest. This way, she takes as many math courses as possible while still fulfilling the requirement that she take at least 35 classes.