Module 5: Practice Problems
Systems of Equations and Inequalities
About These Practice Problems
This set contains 20 comprehensive problems covering all topics in Module 5. Work through each problem carefully, then check your solution. Each problem includes detailed step-by-step explanations.
Problems 1-4: Systems of Linear Equations
Problem 1: Solve by Substitution
Solve the system of equations using the substitution method:
y = 2x - 5
3x + y = 10
Solution:
Step 1: Identify the equation already solved for one variable
The first equation is already solved for y: y = 2x - 5
The first equation is already solved for y: y = 2x - 5
Step 2: Substitute into the second equation
3x + y = 10
3x + (2x - 5) = 10
3x + y = 10
3x + (2x - 5) = 10
Step 3: Solve for x
3x + 2x - 5 = 10
5x - 5 = 10
5x = 15
x = 3
3x + 2x - 5 = 10
5x - 5 = 10
5x = 15
x = 3
Step 4: Substitute x = 3 back into y = 2x - 5
y = 2(3) - 5
y = 6 - 5
y = 1
y = 2(3) - 5
y = 6 - 5
y = 1
Step 5: Check the solution in both equations
First equation: y = 2x - 5 → 1 = 2(3) - 5 → 1 = 1 (Check)
Second equation: 3x + y = 10 → 3(3) + 1 = 10 → 10 = 10 (Check)
First equation: y = 2x - 5 → 1 = 2(3) - 5 → 1 = 1 (Check)
Second equation: 3x + y = 10 → 3(3) + 1 = 10 → 10 = 10 (Check)
Final Answer: (3, 1)
Problem 2: Solve by Elimination
Solve the system of equations using the elimination method:
2x + 3y = 12
5x - 3y = 9
Solution:
Step 1: Observe that the y-coefficients are opposites
First equation: 2x + 3y = 12
Second equation: 5x - 3y = 9
The coefficients of y are 3 and -3, which are opposites.
First equation: 2x + 3y = 12
Second equation: 5x - 3y = 9
The coefficients of y are 3 and -3, which are opposites.
Step 2: Add the equations to eliminate y
(2x + 3y) + (5x - 3y) = 12 + 9
7x = 21
x = 3
(2x + 3y) + (5x - 3y) = 12 + 9
7x = 21
x = 3
Step 3: Substitute x = 3 into either original equation
Using the first equation: 2x + 3y = 12
2(3) + 3y = 12
6 + 3y = 12
3y = 6
y = 2
Using the first equation: 2x + 3y = 12
2(3) + 3y = 12
6 + 3y = 12
3y = 6
y = 2
Step 4: Check the solution
First equation: 2(3) + 3(2) = 6 + 6 = 12 (Check)
Second equation: 5(3) - 3(2) = 15 - 6 = 9 (Check)
First equation: 2(3) + 3(2) = 6 + 6 = 12 (Check)
Second equation: 5(3) - 3(2) = 15 - 6 = 9 (Check)
Final Answer: (3, 2)
Problem 3: Inconsistent System
Solve the system of equations:
2x - y = 4
-4x + 2y = 3
Solution:
Step 1: Use elimination method
Multiply the first equation by 2 to make x-coefficients opposites:
4x - 2y = 8
-4x + 2y = 3
Multiply the first equation by 2 to make x-coefficients opposites:
4x - 2y = 8
-4x + 2y = 3
Step 2: Add the equations
(4x - 2y) + (-4x + 2y) = 8 + 3
0 = 11
(4x - 2y) + (-4x + 2y) = 8 + 3
0 = 11
Step 3: Interpret the result
The statement 0 = 11 is false.
This means the system has no solution.
The statement 0 = 11 is false.
This means the system has no solution.
This is an inconsistent system. The lines are parallel and never intersect.
You can verify by noting that the second equation is -2 times the first equation on the left side, but the right side doesn't follow the same pattern.
You can verify by noting that the second equation is -2 times the first equation on the left side, but the right side doesn't follow the same pattern.
Final Answer: No solution (inconsistent system)
Problem 4: Dependent System
Solve the system of equations:
3x - 6y = 12
-x + 2y = -4
Solution:
Step 1: Use elimination method
Multiply the second equation by 3:
3x - 6y = 12
-3x + 6y = -12
Multiply the second equation by 3:
3x - 6y = 12
-3x + 6y = -12
Step 2: Add the equations
(3x - 6y) + (-3x + 6y) = 12 + (-12)
0 = 0
(3x - 6y) + (-3x + 6y) = 12 + (-12)
0 = 0
Step 3: Interpret the result
The statement 0 = 0 is always true.
This means the system has infinitely many solutions.
The statement 0 = 0 is always true.
This means the system has infinitely many solutions.
Step 4: Express the solution set
Solve the second equation for y:
-x + 2y = -4
2y = x - 4
y = (x - 4)/2 or y = (1/2)x - 2
Solve the second equation for y:
-x + 2y = -4
2y = x - 4
y = (x - 4)/2 or y = (1/2)x - 2
This is a dependent system. The two equations represent the same line.
Any point on the line is a solution.
Any point on the line is a solution.
Final Answer: Infinitely many solutions; y = (1/2)x - 2 (or equivalent)
Problems 5-7: Matrices and Gaussian Elimination
Problem 5: Augmented Matrix and Gaussian Elimination
Solve the system using Gaussian elimination:
x + 2y = 7
3x - y = 5
Solution:
Step 1: Write the augmented matrix
[1 2 | 7]
[3 -1 | 5]
[1 2 | 7]
[3 -1 | 5]
Step 2: Make the first column below the first row all zeros
R2 - 3R1 → R2 (subtract 3 times row 1 from row 2)
[1 2 | 7]
[0 -7 | -16]
R2 - 3R1 → R2 (subtract 3 times row 1 from row 2)
[1 2 | 7]
[0 -7 | -16]
Step 3: Make the leading coefficient in row 2 equal to 1
R2 / (-7) → R2
[1 2 | 7]
[0 1 | 16/7]
R2 / (-7) → R2
[1 2 | 7]
[0 1 | 16/7]
Step 4: Back-substitute to find x
From row 2: y = 16/7
From row 1: x + 2y = 7
x + 2(16/7) = 7
x + 32/7 = 49/7
x = 49/7 - 32/7 = 17/7
From row 2: y = 16/7
From row 1: x + 2y = 7
x + 2(16/7) = 7
x + 32/7 = 49/7
x = 49/7 - 32/7 = 17/7
Step 5: Verify the solution
x + 2y = 17/7 + 2(16/7) = 17/7 + 32/7 = 49/7 = 7 (Check)
3x - y = 3(17/7) - 16/7 = 51/7 - 16/7 = 35/7 = 5 (Check)
x + 2y = 17/7 + 2(16/7) = 17/7 + 32/7 = 49/7 = 7 (Check)
3x - y = 3(17/7) - 16/7 = 51/7 - 16/7 = 35/7 = 5 (Check)
Final Answer: (17/7, 16/7)
Problem 6: 3×3 System Using Gauss-Jordan
Solve the system using Gauss-Jordan elimination:
x + y + z = 6
2x - y + z = 3
x + 2y - z = 2
Solution:
Step 1: Write the augmented matrix
[1 1 1 | 6]
[2 -1 1 | 3]
[1 2 -1 | 2]
[1 1 1 | 6]
[2 -1 1 | 3]
[1 2 -1 | 2]
Step 2: Eliminate below the first pivot
R2 - 2R1 → R2
R3 - R1 → R3
[1 1 1 | 6]
[0 -3 -1 | -9]
[0 1 -2 | -4]
R2 - 2R1 → R2
R3 - R1 → R3
[1 1 1 | 6]
[0 -3 -1 | -9]
[0 1 -2 | -4]
Step 3: Make the second pivot equal to 1
R2 / (-3) → R2
[1 1 1 | 6]
[0 1 1/3 | 3]
[0 1 -2 | -4]
R2 / (-3) → R2
[1 1 1 | 6]
[0 1 1/3 | 3]
[0 1 -2 | -4]
Step 4: Eliminate below the second pivot
R3 - R2 → R3
[1 1 1 | 6]
[0 1 1/3 | 3]
[0 0 -7/3 | -7]
R3 - R2 → R3
[1 1 1 | 6]
[0 1 1/3 | 3]
[0 0 -7/3 | -7]
Step 5: Make the third pivot equal to 1
R3 × (-3/7) → R3
[1 1 1 | 6]
[0 1 1/3 | 3]
[0 0 1 | 3]
R3 × (-3/7) → R3
[1 1 1 | 6]
[0 1 1/3 | 3]
[0 0 1 | 3]
Step 6: Back-substitute using Gauss-Jordan
From row 3: z = 3
From row 2: y + (1/3)z = 3 → y + 1 = 3 → y = 2
From row 1: x + y + z = 6 → x + 2 + 3 = 6 → x = 1
From row 3: z = 3
From row 2: y + (1/3)z = 3 → y + 1 = 3 → y = 2
From row 1: x + y + z = 6 → x + 2 + 3 = 6 → x = 1
Final Answer: (1, 2, 3)
Problem 7: Identify System Type from Row-Echelon Form
Each augmented matrix is in row-echelon form. Determine whether the system is consistent or inconsistent. If consistent, state whether it has one solution or infinitely many solutions.
a) [1 2 | 5]
[0 1 | 3]
b) [1 -1 2 | 4]
[0 1 -3 | 2]
[0 0 0 | 5]
c) [1 3 -2 | 7]
[0 1 4 | -1]
[0 0 0 | 0]
Solution:
Part a)
[1 2 | 5]
[0 1 | 3]
Analysis: The matrix has 2 pivots (leading 1s) in 2 rows.
From row 2: y = 3
From row 1: x + 2(3) = 5 → x = -1
Result: Consistent system with one unique solution: (-1, 3)
[1 2 | 5]
[0 1 | 3]
Analysis: The matrix has 2 pivots (leading 1s) in 2 rows.
From row 2: y = 3
From row 1: x + 2(3) = 5 → x = -1
Result: Consistent system with one unique solution: (-1, 3)
Part b)
[1 -1 2 | 4]
[0 1 -3 | 2]
[0 0 0 | 5]
Analysis: The last row represents 0 = 5, which is false.
Result: Inconsistent system (no solution)
[1 -1 2 | 4]
[0 1 -3 | 2]
[0 0 0 | 5]
Analysis: The last row represents 0 = 5, which is false.
Result: Inconsistent system (no solution)
Part c)
[1 3 -2 | 7]
[0 1 4 | -1]
[0 0 0 | 0]
Analysis: The last row represents 0 = 0, which is always true.
We have 2 pivots but 3 variables, so z is a free variable.
From row 2: y + 4z = -1 → y = -1 - 4z
From row 1: x + 3y - 2z = 7 → x = 7 - 3y + 2z = 7 - 3(-1 - 4z) + 2z = 10 + 14z
Result: Consistent system with infinitely many solutions
[1 3 -2 | 7]
[0 1 4 | -1]
[0 0 0 | 0]
Analysis: The last row represents 0 = 0, which is always true.
We have 2 pivots but 3 variables, so z is a free variable.
From row 2: y + 4z = -1 → y = -1 - 4z
From row 1: x + 3y - 2z = 7 → x = 7 - 3y + 2z = 7 - 3(-1 - 4z) + 2z = 10 + 14z
Result: Consistent system with infinitely many solutions
Final Answers: a) One solution: (-1, 3); b) No solution; c) Infinitely many solutions
Problems 8-11: Nonlinear Systems
Problem 8: Line and Parabola
Solve the system:
y = x2 - 4
y = 2x - 1
Solution:
Step 1: Use substitution
Since both equations are solved for y, set them equal:
x2 - 4 = 2x - 1
Since both equations are solved for y, set them equal:
x2 - 4 = 2x - 1
Step 2: Solve the quadratic equation
x2 - 4 = 2x - 1
x2 - 2x - 4 + 1 = 0
x2 - 2x - 3 = 0
(x - 3)(x + 1) = 0
x = 3 or x = -1
x2 - 4 = 2x - 1
x2 - 2x - 4 + 1 = 0
x2 - 2x - 3 = 0
(x - 3)(x + 1) = 0
x = 3 or x = -1
Step 3: Find corresponding y-values
When x = 3: y = 2(3) - 1 = 5
When x = -1: y = 2(-1) - 1 = -3
When x = 3: y = 2(3) - 1 = 5
When x = -1: y = 2(-1) - 1 = -3
Step 4: Verify both solutions
For (3, 5): 5 = 32 - 4 = 5 (Check) and 5 = 2(3) - 1 = 5 (Check)
For (-1, -3): -3 = (-1)2 - 4 = -3 (Check) and -3 = 2(-1) - 1 = -3 (Check)
For (3, 5): 5 = 32 - 4 = 5 (Check) and 5 = 2(3) - 1 = 5 (Check)
For (-1, -3): -3 = (-1)2 - 4 = -3 (Check) and -3 = 2(-1) - 1 = -3 (Check)
Final Answer: (3, 5) and (-1, -3)
Problem 9: Line and Circle
Solve the system:
x2 + y2 = 25
y = x + 1
Solution:
Step 1: Substitute y = x + 1 into the circle equation
x2 + y2 = 25
x2 + (x + 1)2 = 25
x2 + y2 = 25
x2 + (x + 1)2 = 25
Step 2: Expand and simplify
x2 + (x2 + 2x + 1) = 25
2x2 + 2x + 1 = 25
2x2 + 2x - 24 = 0
x2 + x - 12 = 0
x2 + (x2 + 2x + 1) = 25
2x2 + 2x + 1 = 25
2x2 + 2x - 24 = 0
x2 + x - 12 = 0
Step 3: Factor and solve
(x + 4)(x - 3) = 0
x = -4 or x = 3
(x + 4)(x - 3) = 0
x = -4 or x = 3
Step 4: Find corresponding y-values
When x = -4: y = -4 + 1 = -3
When x = 3: y = 3 + 1 = 4
When x = -4: y = -4 + 1 = -3
When x = 3: y = 3 + 1 = 4
Step 5: Verify both solutions
For (-4, -3): (-4)2 + (-3)2 = 16 + 9 = 25 (Check)
For (3, 4): 32 + 42 = 9 + 16 = 25 (Check)
For (-4, -3): (-4)2 + (-3)2 = 16 + 9 = 25 (Check)
For (3, 4): 32 + 42 = 9 + 16 = 25 (Check)
Final Answer: (-4, -3) and (3, 4)
Problem 10: Two Quadratics
Solve the system:
y = x2
y = 4 - x2
Solution:
Step 1: Set the equations equal
x2 = 4 - x2
x2 = 4 - x2
Step 2: Solve for x
2x2 = 4
x2 = 2
x = ±√2
2x2 = 4
x2 = 2
x = ±√2
Step 3: Find corresponding y-values
When x = √2: y = (√2)2 = 2
When x = -√2: y = (-√2)2 = 2
When x = √2: y = (√2)2 = 2
When x = -√2: y = (-√2)2 = 2
Step 4: Verify both solutions
For (√2, 2): Check first equation: 2 = (√2)2 = 2 (Check)
Check second equation: 2 = 4 - (√2)2 = 4 - 2 = 2 (Check)
For (-√2, 2): Similar verification works
For (√2, 2): Check first equation: 2 = (√2)2 = 2 (Check)
Check second equation: 2 = 4 - (√2)2 = 4 - 2 = 2 (Check)
For (-√2, 2): Similar verification works
Final Answer: (√2, 2) and (-√2, 2)
Problem 11: Circle and Ellipse
Solve the system:
x2 + y2 = 9
x2/4 + y2/9 = 1
Solution:
Step 1: Solve the first equation for x2
x2 = 9 - y2
x2 = 9 - y2
Step 2: Substitute into the second equation
(9 - y2)/4 + y2/9 = 1
Multiply through by 36 to clear denominators:
9(9 - y2) + 4y2 = 36
(9 - y2)/4 + y2/9 = 1
Multiply through by 36 to clear denominators:
9(9 - y2) + 4y2 = 36
Step 3: Simplify and solve for y
81 - 9y2 + 4y2 = 36
81 - 5y2 = 36
-5y2 = -45
y2 = 9
y = ±3
81 - 9y2 + 4y2 = 36
81 - 5y2 = 36
-5y2 = -45
y2 = 9
y = ±3
Step 4: Find corresponding x-values
When y = 3: x2 = 9 - 9 = 0 → x = 0
When y = -3: x2 = 9 - 9 = 0 → x = 0
When y = 3: x2 = 9 - 9 = 0 → x = 0
When y = -3: x2 = 9 - 9 = 0 → x = 0
Step 5: Verify the solutions
For (0, 3): 02 + 32 = 9 (Check) and 0/4 + 9/9 = 1 (Check)
For (0, -3): 02 + (-3)2 = 9 (Check) and 0/4 + 9/9 = 1 (Check)
For (0, 3): 02 + 32 = 9 (Check) and 0/4 + 9/9 = 1 (Check)
For (0, -3): 02 + (-3)2 = 9 (Check) and 0/4 + 9/9 = 1 (Check)
Final Answer: (0, 3) and (0, -3)
Problems 12-15: Systems of Inequalities
Problem 12: Graph a Linear Inequality
Graph the inequality: 2x + y ≤ 6
Determine whether the following points are in the solution region:
a) (0, 0)
b) (4, 0)
c) (2, 2)
Solution:
Step 1: Graph the boundary line
First, graph 2x + y = 6 as a solid line (because of ≤).
Find intercepts:
When x = 0: y = 6 → point (0, 6)
When y = 0: 2x = 6 → x = 3 → point (3, 0)
Draw a solid line through these points.
First, graph 2x + y = 6 as a solid line (because of ≤).
Find intercepts:
When x = 0: y = 6 → point (0, 6)
When y = 0: 2x = 6 → x = 3 → point (3, 0)
Draw a solid line through these points.
Step 2: Determine which side to shade
Test point (0, 0):
2(0) + 0 ≤ 6
0 ≤ 6 (True)
Shade the side containing (0, 0), which is below and to the left of the line.
Test point (0, 0):
2(0) + 0 ≤ 6
0 ≤ 6 (True)
Shade the side containing (0, 0), which is below and to the left of the line.
Step 3: Test the given points
a) (0, 0): 2(0) + 0 = 0 ≤ 6 (True - in the solution region)
b) (4, 0): 2(4) + 0 = 8 ≤ 6 (False - NOT in the solution region)
c) (2, 2): 2(2) + 2 = 6 ≤ 6 (True - on the boundary, included)
a) (0, 0): 2(0) + 0 = 0 ≤ 6 (True - in the solution region)
b) (4, 0): 2(4) + 0 = 8 ≤ 6 (False - NOT in the solution region)
c) (2, 2): 2(2) + 2 = 6 ≤ 6 (True - on the boundary, included)
Final Answer: Graph shows region below/left of line 2x + y = 6 (solid line). Points (0,0) and (2,2) are in the solution region; (4,0) is not.
Problem 13: System of Inequalities with Vertices
Graph the system of inequalities and identify the vertices of the feasible region:
x + y ≤ 6
x - y ≤ 2
x ≥ 0
y ≥ 0
Solution:
Step 1: Graph each boundary line
Line 1: x + y = 6 (intercepts at (6, 0) and (0, 6))
Line 2: x - y = 2 (intercepts at (2, 0) and (0, -2))
Line 3: x = 0 (y-axis)
Line 4: y = 0 (x-axis)
Line 1: x + y = 6 (intercepts at (6, 0) and (0, 6))
Line 2: x - y = 2 (intercepts at (2, 0) and (0, -2))
Line 3: x = 0 (y-axis)
Line 4: y = 0 (x-axis)
Step 2: Determine the feasible region
The feasible region is where all inequalities are satisfied simultaneously.
x + y ≤ 6: region below the line
x - y ≤ 2: region above the line
x ≥ 0: region to the right of y-axis
y ≥ 0: region above x-axis
The feasible region is where all inequalities are satisfied simultaneously.
x + y ≤ 6: region below the line
x - y ≤ 2: region above the line
x ≥ 0: region to the right of y-axis
y ≥ 0: region above x-axis
Step 3: Find the vertices (corner points)
Vertex 1: Intersection of x = 0 and y = 0 → (0, 0)
Vertex 2: Intersection of x = 0 and x + y = 6 → (0, 6)
Vertex 3: Intersection of x + y = 6 and x - y = 2
Add equations: 2x = 8 → x = 4
Substitute: 4 + y = 6 → y = 2 → (4, 2)
Vertex 4: Intersection of x - y = 2 and y = 0 → (2, 0)
Vertex 1: Intersection of x = 0 and y = 0 → (0, 0)
Vertex 2: Intersection of x = 0 and x + y = 6 → (0, 6)
Vertex 3: Intersection of x + y = 6 and x - y = 2
Add equations: 2x = 8 → x = 4
Substitute: 4 + y = 6 → y = 2 → (4, 2)
Vertex 4: Intersection of x - y = 2 and y = 0 → (2, 0)
Step 4: Verify vertices are in the feasible region
All four vertices satisfy all four inequalities.
All four vertices satisfy all four inequalities.
Final Answer: Vertices of the feasible region are (0, 0), (0, 6), (4, 2), and (2, 0)
Problem 14: Determine if Point Satisfies System
Determine whether the point (3, 1) satisfies the system of inequalities:
2x - y > 4
x + 2y < 8
x ≥ 0
y ≥ 0
Solution:
Test the point (3, 1) in each inequality:
Inequality 1: 2x - y > 4
2(3) - 1 > 4
6 - 1 > 4
5 > 4 (True)
2(3) - 1 > 4
6 - 1 > 4
5 > 4 (True)
Inequality 2: x + 2y < 8
3 + 2(1) < 8
3 + 2 < 8
5 < 8 (True)
3 + 2(1) < 8
3 + 2 < 8
5 < 8 (True)
Inequality 3: x ≥ 0
3 ≥ 0 (True)
3 ≥ 0 (True)
Inequality 4: y ≥ 0
1 ≥ 0 (True)
1 ≥ 0 (True)
Conclusion:
Since the point (3, 1) satisfies all four inequalities, it is in the solution region.
Since the point (3, 1) satisfies all four inequalities, it is in the solution region.
Final Answer: Yes, (3, 1) satisfies the system
Problem 15: Write System from Graph
Write a system of inequalities that describes a triangular region with vertices at (0, 0), (5, 0), and (0, 4).
Solution:
Step 1: Identify the boundary lines
The triangle has three sides:
Side 1: From (0, 0) to (5, 0) along the x-axis
Side 2: From (0, 0) to (0, 4) along the y-axis
Side 3: From (5, 0) to (0, 4), which is the hypotenuse
The triangle has three sides:
Side 1: From (0, 0) to (5, 0) along the x-axis
Side 2: From (0, 0) to (0, 4) along the y-axis
Side 3: From (5, 0) to (0, 4), which is the hypotenuse
Step 2: Write equations for the boundary lines
Side 1: y = 0 (x-axis)
Side 2: x = 0 (y-axis)
Side 3: Find the equation of the line through (5, 0) and (0, 4)
Slope: m = (4 - 0)/(0 - 5) = -4/5
Using point-slope form with (0, 4): y - 4 = (-4/5)(x - 0)
y = -4x/5 + 4
Or in standard form: 4x + 5y = 20
Side 1: y = 0 (x-axis)
Side 2: x = 0 (y-axis)
Side 3: Find the equation of the line through (5, 0) and (0, 4)
Slope: m = (4 - 0)/(0 - 5) = -4/5
Using point-slope form with (0, 4): y - 4 = (-4/5)(x - 0)
y = -4x/5 + 4
Or in standard form: 4x + 5y = 20
Step 3: Determine inequality directions
The region is inside the triangle, so:
x ≥ 0 (to the right of the y-axis)
y ≥ 0 (above the x-axis)
For the hypotenuse, test point (0, 0):
4(0) + 5(0) = 0 ≤ 20 (True)
So: 4x + 5y ≤ 20
The region is inside the triangle, so:
x ≥ 0 (to the right of the y-axis)
y ≥ 0 (above the x-axis)
For the hypotenuse, test point (0, 0):
4(0) + 5(0) = 0 ≤ 20 (True)
So: 4x + 5y ≤ 20
Final Answer: x ≥ 0, y ≥ 0, and 4x + 5y ≤ 20
Problems 16-20: Linear Programming Applications
Problem 16: Set Up Constraints and Objective Function
A furniture maker produces chairs and tables. Each chair requires 2 hours of labor and 4 board feet of wood. Each table requires 3 hours of labor and 6 board feet of wood. The maker has 60 hours of labor and 120 board feet of wood available. Chairs sell for $50 profit and tables for $80 profit.
a) Define variables and write the constraints
b) Write the objective function to maximize profit
Solution:
Part a) Define variables:
Let x = number of chairs produced
Let y = number of tables produced
Let x = number of chairs produced
Let y = number of tables produced
Write constraints:
Labor constraint: 2x + 3y ≤ 60
(Each chair takes 2 hours, each table takes 3 hours, total 60 hours available)
Wood constraint: 4x + 6y ≤ 120
(Each chair needs 4 board feet, each table needs 6, total 120 available)
Non-negativity constraints: x ≥ 0, y ≥ 0
(Cannot produce negative furniture)
Labor constraint: 2x + 3y ≤ 60
(Each chair takes 2 hours, each table takes 3 hours, total 60 hours available)
Wood constraint: 4x + 6y ≤ 120
(Each chair needs 4 board feet, each table needs 6, total 120 available)
Non-negativity constraints: x ≥ 0, y ≥ 0
(Cannot produce negative furniture)
Part b) Objective function:
Maximize P = 50x + 80y
(Profit is $50 per chair plus $80 per table)
Maximize P = 50x + 80y
(Profit is $50 per chair plus $80 per table)
Final Answer: Constraints: 2x + 3y ≤ 60, 4x + 6y ≤ 120, x ≥ 0, y ≥ 0; Objective: Maximize P = 50x + 80y
Problem 17: Solve Linear Programming Problem Graphically
Using the furniture problem from Problem 16, find the optimal production plan that maximizes profit.
Constraints: 2x + 3y ≤ 60, 4x + 6y ≤ 120, x ≥ 0, y ≥ 0
Objective: Maximize P = 50x + 80y
Solution:
Step 1: Simplify constraints if possible
4x + 6y ≤ 120 can be simplified by dividing by 2:
2x + 3y ≤ 60
This is the same as the first constraint! So we really only have one main constraint.
4x + 6y ≤ 120 can be simplified by dividing by 2:
2x + 3y ≤ 60
This is the same as the first constraint! So we really only have one main constraint.
Step 2: Find the vertices of the feasible region
Vertex 1: (0, 0)
Vertex 2: Intersection of x = 0 and 2x + 3y = 60 → (0, 20)
Vertex 3: Intersection of 2x + 3y = 60 and y = 0 → (30, 0)
Vertex 1: (0, 0)
Vertex 2: Intersection of x = 0 and 2x + 3y = 60 → (0, 20)
Vertex 3: Intersection of 2x + 3y = 60 and y = 0 → (30, 0)
Step 3: Evaluate the objective function at each vertex
At (0, 0): P = 50(0) + 80(0) = $0
At (0, 20): P = 50(0) + 80(20) = $1,600
At (30, 0): P = 50(30) + 80(0) = $1,500
At (0, 0): P = 50(0) + 80(0) = $0
At (0, 20): P = 50(0) + 80(20) = $1,600
At (30, 0): P = 50(30) + 80(0) = $1,500
Step 4: Identify the maximum
The maximum profit is $1,600, which occurs at (0, 20).
The maximum profit is $1,600, which occurs at (0, 20).
Final Answer: Produce 0 chairs and 20 tables for maximum profit of $1,600
Problem 18: Manufacturing Problem
A manufacturer produces two types of products, A and B. Product A requires 1 hour on Machine 1 and 2 hours on Machine 2. Product B requires 2 hours on Machine 1 and 1 hour on Machine 2. Machine 1 is available for 40 hours per week and Machine 2 for 50 hours per week. Product A yields a profit of $30 and Product B yields $40. How many of each product should be made to maximize profit?
Solution:
Step 1: Define variables and constraints
Let x = number of Product A
Let y = number of Product B
Machine 1 constraint: x + 2y ≤ 40
Machine 2 constraint: 2x + y ≤ 50
Non-negativity: x ≥ 0, y ≥ 0
Objective: Maximize P = 30x + 40y
Let x = number of Product A
Let y = number of Product B
Machine 1 constraint: x + 2y ≤ 40
Machine 2 constraint: 2x + y ≤ 50
Non-negativity: x ≥ 0, y ≥ 0
Objective: Maximize P = 30x + 40y
Step 2: Find vertices of feasible region
Vertex 1: (0, 0)
Vertex 2: Intersection of x = 0 and x + 2y = 40 → (0, 20)
Vertex 3: Intersection of x + 2y = 40 and 2x + y = 50
From first equation: x = 40 - 2y
Substitute: 2(40 - 2y) + y = 50
80 - 4y + y = 50
-3y = -30 → y = 10
x = 40 - 2(10) = 20 → (20, 10)
Vertex 4: Intersection of 2x + y = 50 and y = 0 → (25, 0)
Vertex 1: (0, 0)
Vertex 2: Intersection of x = 0 and x + 2y = 40 → (0, 20)
Vertex 3: Intersection of x + 2y = 40 and 2x + y = 50
From first equation: x = 40 - 2y
Substitute: 2(40 - 2y) + y = 50
80 - 4y + y = 50
-3y = -30 → y = 10
x = 40 - 2(10) = 20 → (20, 10)
Vertex 4: Intersection of 2x + y = 50 and y = 0 → (25, 0)
Step 3: Evaluate objective function
At (0, 0): P = 30(0) + 40(0) = $0
At (0, 20): P = 30(0) + 40(20) = $800
At (20, 10): P = 30(20) + 40(10) = $600 + $400 = $1,000
At (25, 0): P = 30(25) + 40(0) = $750
At (0, 0): P = 30(0) + 40(0) = $0
At (0, 20): P = 30(0) + 40(20) = $800
At (20, 10): P = 30(20) + 40(10) = $600 + $400 = $1,000
At (25, 0): P = 30(25) + 40(0) = $750
Final Answer: Produce 20 units of Product A and 10 units of Product B for maximum profit of $1,000
Problem 19: Diet/Nutrition Problem
A nutritionist plans a meal using Food X and Food Y. Food X contains 2 grams of protein and 3 grams of carbohydrates per serving. Food Y contains 3 grams of protein and 2 grams of carbohydrates per serving. The meal must provide at least 18 grams of protein and at least 18 grams of carbohydrates. Food X costs $0.50 per serving and Food Y costs $0.60 per serving. Find the minimum cost meal that meets the nutritional requirements.
Solution:
Step 1: Define variables and constraints
Let x = servings of Food X
Let y = servings of Food Y
Protein constraint: 2x + 3y ≥ 18
Carbohydrate constraint: 3x + 2y ≥ 18
Non-negativity: x ≥ 0, y ≥ 0
Objective: Minimize C = 0.50x + 0.60y
Let x = servings of Food X
Let y = servings of Food Y
Protein constraint: 2x + 3y ≥ 18
Carbohydrate constraint: 3x + 2y ≥ 18
Non-negativity: x ≥ 0, y ≥ 0
Objective: Minimize C = 0.50x + 0.60y
Step 2: Find vertices of feasible region
Vertex 1: Intersection of 2x + 3y = 18 and x = 0 → (0, 6)
Vertex 2: Intersection of 2x + 3y = 18 and 3x + 2y = 18
Multiply first by 3: 6x + 9y = 54
Multiply second by 2: 6x + 4y = 36
Subtract: 5y = 18 → y = 18/5 = 3.6
2x + 3(3.6) = 18 → 2x = 7.2 → x = 3.6 → (3.6, 3.6)
Vertex 3: Intersection of 3x + 2y = 18 and y = 0 → (6, 0)
Vertex 1: Intersection of 2x + 3y = 18 and x = 0 → (0, 6)
Vertex 2: Intersection of 2x + 3y = 18 and 3x + 2y = 18
Multiply first by 3: 6x + 9y = 54
Multiply second by 2: 6x + 4y = 36
Subtract: 5y = 18 → y = 18/5 = 3.6
2x + 3(3.6) = 18 → 2x = 7.2 → x = 3.6 → (3.6, 3.6)
Vertex 3: Intersection of 3x + 2y = 18 and y = 0 → (6, 0)
Step 3: Evaluate cost at each vertex
At (0, 6): C = 0.50(0) + 0.60(6) = $3.60
At (3.6, 3.6): C = 0.50(3.6) + 0.60(3.6) = $1.80 + $2.16 = $3.96
At (6, 0): C = 0.50(6) + 0.60(0) = $3.00
At (0, 6): C = 0.50(0) + 0.60(6) = $3.60
At (3.6, 3.6): C = 0.50(3.6) + 0.60(3.6) = $1.80 + $2.16 = $3.96
At (6, 0): C = 0.50(6) + 0.60(0) = $3.00
Step 4: Identify the minimum
The minimum cost is $3.00, which occurs at (6, 0).
The minimum cost is $3.00, which occurs at (6, 0).
Final Answer: Use 6 servings of Food X and 0 servings of Food Y for minimum cost of $3.00
Problem 20: Mixed Application
A farmer has 100 acres to plant corn and soybeans. Corn requires 3 hours of labor per acre and soybeans require 2 hours per acre. The farmer has 240 hours of labor available. Corn yields a profit of $200 per acre and soybeans yield $150 per acre. Additionally, market conditions require that at least 20 acres of corn be planted. How many acres of each crop should be planted to maximize profit?
Solution:
Step 1: Define variables and constraints
Let x = acres of corn
Let y = acres of soybeans
Land constraint: x + y ≤ 100
Labor constraint: 3x + 2y ≤ 240
Minimum corn: x ≥ 20
Non-negativity: x ≥ 0, y ≥ 0 (redundant with x ≥ 20)
Objective: Maximize P = 200x + 150y
Let x = acres of corn
Let y = acres of soybeans
Land constraint: x + y ≤ 100
Labor constraint: 3x + 2y ≤ 240
Minimum corn: x ≥ 20
Non-negativity: x ≥ 0, y ≥ 0 (redundant with x ≥ 20)
Objective: Maximize P = 200x + 150y
Step 2: Find vertices of feasible region
Vertex 1: Intersection of x = 20 and y = 0 → (20, 0)
Vertex 2: Intersection of x = 20 and x + y = 100 → (20, 80)
Vertex 3: Intersection of x + y = 100 and 3x + 2y = 240
From first: y = 100 - x
Substitute: 3x + 2(100 - x) = 240
3x + 200 - 2x = 240
x = 40 → y = 60 → (40, 60)
Vertex 4: Intersection of 3x + 2y = 240 and y = 0 → (80, 0)
Vertex 1: Intersection of x = 20 and y = 0 → (20, 0)
Vertex 2: Intersection of x = 20 and x + y = 100 → (20, 80)
Vertex 3: Intersection of x + y = 100 and 3x + 2y = 240
From first: y = 100 - x
Substitute: 3x + 2(100 - x) = 240
3x + 200 - 2x = 240
x = 40 → y = 60 → (40, 60)
Vertex 4: Intersection of 3x + 2y = 240 and y = 0 → (80, 0)
Step 3: Check which vertices are in the feasible region
(20, 0): Satisfies all constraints (Check)
(20, 80): Check 3x + 2y ≤ 240: 3(20) + 2(80) = 220 ≤ 240 (Check)
(40, 60): Satisfies all constraints (Check)
(80, 0): Check x ≥ 20: 80 ≥ 20 (Check)
(20, 0): Satisfies all constraints (Check)
(20, 80): Check 3x + 2y ≤ 240: 3(20) + 2(80) = 220 ≤ 240 (Check)
(40, 60): Satisfies all constraints (Check)
(80, 0): Check x ≥ 20: 80 ≥ 20 (Check)
Step 4: Evaluate profit at each vertex
At (20, 0): P = 200(20) + 150(0) = $4,000
At (20, 80): P = 200(20) + 150(80) = $4,000 + $12,000 = $16,000
At (40, 60): P = 200(40) + 150(60) = $8,000 + $9,000 = $17,000
At (80, 0): P = 200(80) + 150(0) = $16,000
At (20, 0): P = 200(20) + 150(0) = $4,000
At (20, 80): P = 200(20) + 150(80) = $4,000 + $12,000 = $16,000
At (40, 60): P = 200(40) + 150(60) = $8,000 + $9,000 = $17,000
At (80, 0): P = 200(80) + 150(0) = $16,000
Final Answer: Plant 40 acres of corn and 60 acres of soybeans for maximum profit of $17,000
Ready to Test Your Knowledge?
Now that you have practiced these problems, take the Module 5 Quiz to demonstrate your mastery!