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Published by , 2018-09-21 09:09:32

W3.3_4.1ACompleted

W3.3_4.1ACompleted

MA 162 - Section 3.3 and 4.1A

Completed

Section 3.3

1. Write out in a numbered list the steps involved in solving a linear programming problem
graphically.

Solution:

(a) Graph the feasible region.

(b) Find the coordinates of all corner points.

(c) Evaluate the objective function at each corner point.

(d) Find the corner point (or points) that maximize (or minimize) the objective function.
Remember that if the maximium occurs at two adjacent corner points, the objective
function is maximized anywhere on the line segment between them.

2. Solve the following linear programming problems graphically.

(a) (b)

Maximize P = 4x + 5y Minimize C = 5x + 3y
subject to x + 2y ≤ 10 subject to 5x + 3y ≥ 30

5x + 3y ≤ 30 x − 3y ≤ 0
x ≥ 0, y ≥ 0 x≥2

Solution: Having graphed the inequalities, we see that we have corner points at the
intersections of: x = 0 and x + 2y = 10; x + 2y = 10 and 5x + 3y = 30; 5x + 3y = 30 and
y = 0; x = 0 and y = 0.
Solving the first system x = 0, x + 2y = 10 gives (0, 5) as the intersection point. The
system x + 2y = 10, 5x + 3y = 30 gives the intersection point (30/7, 20/7). The system
5x + 3y = 30, y = 0 gives (6, 0). Finally, the last system gives (0, 0).

1

y

6

5 x + 2y = 10

4

3

2

1 5x + 3y = 30
0x

-1 0 1 2 3 4 5 6

(a) -1
Having found the coordinates of the corner points, we plug each pair into our objective
function to find the largest value:

4 · 0 + 5 · 5 =25

4 · 30 +5 · 20 220 = 31.43
=
7 77

4 · 6 + 5 · 0 =24

4 · 0 + 5 · 0 =0

We conclude that out objective function takes a maximum value of 220 when x = 30 and
7 7
20
y = 7 .

y

7

6

5

45x + 3y = 30

3

2

x − 3y = 0

1

0x

-1 0 1 2 3 4 5 6 7

(b) -1
In this case, to find the corner points, we need to solve the systems x = 2, 5x + 3y = 30
and 5x + 3y = 30, x − 3y = 0. The first system give us the corner point (2, 20/7) and the
second gives (5, 5/3).
Trying each of these in our objective function reveals that both produce the value of 30.

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Thus, we conclude that our objective function attains a minimum value of 30 at any point
on the line segment from (2, 20/3) to (5, 5/3).

3. National Business Machines manufactures two models of portable printers: A and B. Each
model A costs $100, and each model B costs $150. The profits are $30 for each model A and
$40 for each model B portable printer. If the total number of portable printers demanded per
month does not exceed 2500 and the company has earmarked no more than $600, 000/month
for manufacturing costs, how many units of each model should Nation make each month to
maximize its monthly profit? Solve this problem graphically and interpret your solution.

Solution: Since the total number of portable printers demanded does not exceed 2500,
we have A + B ≤ 2500. Since the company can spend at most 600, 000/month, we have
100A + 150B ≤ 600, 000 since printers A and B costs $100 and $150 respectively. We also
have the real world restraints: A ≥ 0 and B ≥ 0. The objective function is P = 30A+40B.
This gives us the linear programming problem

Maximize P = 30A + 40B
subject to A + B ≤ 2500

100A + 150B ≤ 600, 000
A ≥ 0, B ≥ 0

Graphing the feasible region results in the following:
y

(0, 4000) 100A + 150B = 600, 000
(0, 2500)

A + B = 2500

(2500, 0) x
(6000, 0)

Checking each of the three corner points in our objective function shows that the largest

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value is obtained when A = 0, B = 2500. This means they should make 2500 of the type
B printers and none of the type A printers.

Section 4.1A

1. Identify the pivot column and row in the following simplex tableau:

x yzuvwP C
2 1 1 1 0 0 0 14
4 2 3 0 1 0 0 28
2 5 5 0 0 1 0 30
−1 −2 1 0 0 0 1 0

Solution: Since −2 has the largest absolute value in the last row, the second column is

the pivot column. Then from the first row, 30 is the smallest ratio from the pivot column,
5

so the third row will be the pivot row.

2. Solve the following linear programming problems using the simplex method:

(a)

Maximize P = 3x + 4y
subject to x + y ≤ 4

2x + y ≤ 5
x ≥ 0, y ≥ 0

Solution: First we write down the initial simplex tableau with columns x, y, u, v, P

and constants:  1 1 1 0 0 4

 2 1 0 1 0 5

−3 −4 0 0 1 0

The pivot column is the second column, since −4 has the greatest absolute value.

Then the row with the smallest ratio is the first, since 4 < 5 . So we perform row
1 1
1

operations to transform the second column into 0:

0

1 1 1 0 0 4
−R−2−:R−2−−−R→1  1 0 −1 1 0 1 
−R−3−:R−3+−4−R→1 1 0 4 0 1 16

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Then since there are no negative numbers in the last row, we have arrived at a
maximal solution. The variables not in unit form are set equal to 0, so x = u = 0.
Then y = 4, v = 1, and P = 16.

(b)

Maximize P = 4x + 6y
subject to 3x + 2y ≤ 32

2x + 2y ≤ 23
x + 2y ≤ 16
x ≥ 0, y ≥ 0

Solution: First we write down the initial simplex tableau with columns x, y, u, v, w, P

and constants:  3 2 1 0 0 0 32 

 2 2 0 1 0 0 23 

 1 2 0 0 1 0 16 
 

−4 −6 0 0 0 1 0

The pivot column is the second column, since −6 has the greatest absolute value.

Then the row with the smallest ratio is the third, since 16 < 23 < 32 . So we perform
2 2 1
0

row operations to transform the second column into 01, starting with R3 : 1 R3.
2

0

Then we have:

 3 1 1 0 0 0 24  −R−1−:R−1−−2−R→3  2 0 1 0 −1 0 16 

 2 1 0 1 0 0 18  −R−2−:R−2−−2−R→3  1 0 0 1 −1 0 7 
1   
2 1 0 0 1 0 8   1 1 0 0 1 0 8 
2 2 2

−4 −6 0 0 0 1 0 −R−4−:R−4+−6−R→3 −1 0 0 0 3 1 48

There is still a negative number in the last row, so we have not yet arrived at an

optimal solution. Since there is only one negative number, and it is in the first column,

the first column must be our pivot column. Then the pivot row is the second row

since 7 < 16 < 8 . So we perform row operations to transform the second column
1 2
1

2

0

into 01:

0

−R−1−:R−1−−2−R→2  0 0 1 −2 1 0 2 

−R−3−:R−3−−−12−R→2 1 0 0 1 −1 0 7 

 0 1 0 −1 10 9
 2 2

−R−4−:R−4−+−R→2 0 0 0 1 2 1 55

Page 5

Then since there are no negative numbers in the last row, we have arrived at a

maximal solution. The variables not in unit form are set equal to 0, so v = w = 0.

Then x = 7, y = 9 , u = 2, and P = 55.
2

(c)

Maximize P = 4x + 5y + 6z
subject to 2x + 3y + z ≤ 900

3x + y + z ≤ 350
4x + 2y + z ≤ 400
x ≥ 0, y ≥ 0, z ≥ 0

Solution: First we write down the initial simplex tableau with
columns for x, y, z, u, v, w, P and constants:

 2 3 1 1 0 0 0 900 

 3 1 1 0 1 0 0 350 

 4 2 1 0 0 1 0 400 
 

−4 −5 −6 0 0 0 1 0

The pivot column is the column with the most negative entry in the last row, namely
column 3, the z column. We now look at ratios of the constant column with the z
column. The ratios we get are 900/1, 350/1, and 400/1. The smallest of which is the
second ratio. Thus the pivot is in row 2, column 3. We now perform row operations

0
to get the third column to be 10.

0

−R−1−:R−1−−−R→2  −1 2 0 1 −1 0 0 550 

 3 1 1 0 1 0 0 350 

−R−3−:R−3−−−R→2  1 1 0 0 −1 1 0 50 
 

−R−4−:R−4+−6−R→2 14 1 0 0 6 0 1 2100

Since there are no longer any negative entries in the last row, we are done with the
iteration. Now, the value of the variable heading each unit column (a column of
one 1 and the rest 0’s) is given by the entry lying in the column of constants in the
row containing the 1. Thus we have z = 350, u = 550, w = 50 and P = 2100. The
variables heading columns not in unit form are assigned value zero. Thus we have
x = y = v = 0.

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