Finite Mathematics Quiz: Graphing Feasible Regions

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Graphing Feasible RegionsQuestion 1 of 12

The feasible region for a system is bounded by $y \leq -2x + 10$ , $y \leq x + 1$ , and $y \geq 0$ . If we want to find the corner point that maximizes the expression $3x + 2y$ , which point should we evaluate?

The intersection of

y = -2x + 10

and

y = 0

The intersection of

y = x + 1

and

y = 0

The intersection of

y = -2x + 10

and

y = x + 1

We must evaluate all corner points to determine the maximum

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Finite Mathematics Quiz

Finite Mathematics Quiz: Graphing Feasible Regions

Practice Graphing Feasible Regions in Finite Mathematics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Graphing Feasible Regions, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The intersection of $y = -2x + 10$ and $y = 0$
The intersection of $y = x + 1$ and $y = 0$
The intersection of $y = -2x + 10$ and $y = x + 1$
We must evaluate all corner points to determine the maximum (correct answer)

Explanation: While the question asks which point we should evaluate to find the maximum, the fundamental principle of linear programming is that the optimal value of a linear objective function over a feasible region occurs at a corner point, but we cannot determine which corner point is optimal without evaluating the objective function at all corner points. The three corner points are: (1) y = -2x + 10 and y = 0 gives (5,0), (2) y = x + 1 and y = 0 gives (-1,0), but x = -1 may not be in the feasible region depending on other constraints, (3) y = -2x + 10 and y = x + 1 gives -2x + 10 = x + 1, so 9 = 3x, giving x = 3 and y = 4, so (3,4). We must evaluate 3x + 2y at all feasible corner points to find the maximum. Choices A, B, and C each suggest evaluating only one point, which is incorrect methodology.

Question 2

A feasible region is defined by the system $x \ge 0$ , $y \ge 0$ , $2x+y \le 10$ , and $x+y \le 7$ . The corner point $P$ is formed by the intersection of the lines $2x+y=10$ and $x+y=7$ . If the constraint $x+y \le 7$ is changed to $x+y \le 8$ , what are the coordinates of the new corner point $P'$ formed by the intersection of the corresponding boundary lines?

$P'$ is at $(3,4)$
$P'$ is at $(2,6)$ (correct answer)
$P'$ is at $(4,3)$
$P'$ is at $(1,8)$

Explanation: The original corner point

P

is the solution to the system

2x+y=10

and

x+y=7

. Subtracting the second equation from the first gives

x=3

, and substituting back gives

y=4

. So,

P=(3,4)

. The constraint

x+y \le 7

is changed to

x+y \le 8

. The new corner point

P'

is formed by the intersection of the new boundary line

x+y=8

and the unchanged boundary line

2x+y=10

. To find

P'

, we solve the new system. Subtracting

x+y=8

from

2x+y=10

gives

x=2

. Substituting

x=2

into

x+y=8

gives

2+y=8

, so

y=6

. The new corner point is

P'=(2,6)

Question 3

A company manufactures two types of widgets: standard and deluxe. Each standard widget requires 2 hours of assembly and 1 hour of finishing. Each deluxe widget requires 2 hours of assembly and 3 hours of finishing. The company has a maximum of 160 hours of assembly time and 150 hours of finishing time available per week.

Let $x$ be the number of standard widgets and $y$ be the number of deluxe widgets produced. Which of the following is NOT a corner point of the feasible region for production?

(0, 50)
(45, 35)
(80, 0)
(0, 80) (correct answer)

Explanation: First, formulate the constraints based on the passage. The non-negativity constraints are

x \ge 0

and

y \ge 0

. The assembly constraint is

2x + 2y \le 160

, which simplifies to

x + y \le 80

. The finishing constraint is

x + 3y \le 150

. The corner points of the feasible region are (0,0), the valid intercepts, and the intersection of the boundary lines. The y-intercept is the smaller of

y=80

(from

x+y=80

) and

y=50

(from

x+3y=150

), so (0,50) is a vertex. The x-intercept is the smaller of

x=80

and

x=150

, so (80,0) is a vertex. The intersection of

x+y=80

and

x+3y=150

is found by solving the system. From

x=80-y

, we get

(80-y)+3y=150 \implies 2y=70 \implies y=35

. Then

x=80-35=45

. The intersection is (45,35). So, the corner points are (0,0), (80,0), (0,50), and (45,35). The point (0,80) is the y-intercept of

x+y=80

, but it violates the finishing constraint (

0 + 3(80) = 240 \not\le 150

), so it is not a corner point of the feasible region.

Question 4

A factory produces chairs ( $x$ ) and tables ( $y$ ). Production is subject to a materials constraint, $2x + 3y \le 180$ , and a labor constraint, $2x + y \le 100$ . Standard non-negativity constraints $x \ge 0$ and $y \ge 0$ also apply.

Which ordered pair represents the corner point of the feasible region corresponding to the intersection of the material and labor constraint boundaries?

(30, 40) (correct answer)
(0, 60)
(0, 100)
(50, 0)

Explanation: Linear programming problems like this involve finding the feasible region defined by constraint inequalities, and the optimal solutions always occur at corner points (vertices) of this region. When asked for the intersection of two constraint boundaries, you're looking for where the lines formed by converting the inequalities to equations intersect. To find where the material and labor constraint boundaries intersect, you need to solve the system of equations formed by treating both constraints as equalities:

2x + 3y = 180

(material constraint) and

2x + y = 100

(labor constraint). From the second equation,

y = 100 - 2x

. Substituting into the first equation:

2x + 3(100 - 2x) = 180

, which gives

2x + 300 - 6x = 180

, so

-4x = -120

and

x = 30

. Then

y = 100 - 2(30) = 40

. Therefore, the intersection point is (30, 40). Choice A) (30, 40) is correct as shown above. Choice B) (0, 60) represents where the material constraint intersects the y-axis, not the intersection of both constraints. Choice C) (0, 100) is where the labor constraint intersects the y-axis. Choice D) (50, 0) is where the labor constraint intersects the x-axis. When finding constraint intersections, always convert both inequalities to equations and solve the resulting system algebraically. Don't confuse intersection points with axis intercepts of individual constraints.

Question 5

Determine the set of corner points for the feasible region defined by the system of inequalities: $x \ge 0$ , $y \ge 0$ , $x + y \ge 5$ , and $2x + y \le 4$ .

{(-1, 6)}
The empty set (correct answer)
{(5, 0), (0, 5)}
{(2, 0), (0, 4)}

Explanation: When you encounter a linear programming problem asking for corner points of a feasible region, you need to find where the boundary lines intersect within the region that satisfies all constraints simultaneously. Let's analyze this system:

x \ge 0

y \ge 0

x + y \ge 5

, and

2x + y \le 4

. The first two constraints restrict us to the first quadrant. The third constraint requires points to be on or above the line

x + y = 5

, while the fourth requires points to be on or below the line

2x + y = 4

. Here's the crucial insight: these constraints are contradictory. For any point to satisfy

x + y \ge 5

with

x, y \ge 0

, the minimum value of

x + y

is 5. But if

x + y \ge 5

and we need

2x + y \le 4

, then

2x + y = x + (x + y) \ge x + 5

. Since

x \ge 0

, we have

2x + y \ge 5

, which contradicts

2x + y \le 4

. Therefore, no points can satisfy all four constraints simultaneously, making the feasible region empty. Answer choice A gives

(-1, 6)

, but this violates

x \ge 0

. Choice C lists

(5, 0)

and

(0, 5)

, but

(5, 0)

gives

2x + y = 10 > 4

, violating the fourth constraint. Choice D suggests

(2, 0)

and

(0, 4)

, but

(2, 0)

gives

x + y = 2 < 5

, violating the third constraint. Always check whether the constraints can be satisfied simultaneously before looking for intersection points—contradictory systems yield empty feasible regions.

Question 6

A feasible region has corner points at $(1,2)$ , $(4,1)$ , $(3,5)$ , and $(0,6)$ . If we apply the transformation that replaces each point $(x,y)$ with $(2x, y-1)$ , what is the area of the transformed feasible region compared to the original?

The area is halved
The area remains the same
The area is doubled (correct answer)
The area is quadrupled

Explanation: The transformation (x,y) → (2x, y-1) can be represented by the matrix multiplication with scaling factor 2 in the x-direction and translation by -1 in the y-direction. For area calculations, only the scaling matters (translation doesn't change area). The transformation matrix for the scaling part is [[2,0],[0,1]], which has determinant 2. The area of any region is multiplied by the absolute value of the determinant of the transformation matrix. Since |det| = |2| = 2, the area is doubled. The transformed corner points are: (1,2)→(2,1), (4,1)→(8,0), (3,5)→(6,4), (0,6)→(0,5). Choice A (halved) would occur with determinant 1/2. Choice B (same) would occur with determinant 1. Choice D (quadrupled) would occur with determinant 4.

Question 7

Consider the constraint set: $x + y \geq 3$ , $2x - y \leq 6$ , $x \leq 5$ , and $y \leq 4$ . The feasible region is unbounded in which direction?

The feasible region is bounded in all directions (correct answer)
The feasible region extends infinitely in the positive $x$ direction only
The feasible region extends infinitely in the positive $y$ direction only
The feasible region extends infinitely in both positive $x$ and $y$ directions

Explanation: To determine boundedness, we examine the constraints: x + y ≥ 3 creates a region above and to the right of the line x + y = 3; 2x - y ≤ 6 creates a region below and to the left of the line 2x - y = 6; x ≤ 5 bounds the region from the right; y ≤ 4 bounds the region from above. The intersection of these four half-planes creates a bounded region because we have upper bounds on both x (x ≤ 5) and y (y ≤ 4), and the other two constraints further restrict the feasible region without extending it to infinity. The corner points can be found by solving the boundary intersections, and all will have finite coordinates, confirming the region is bounded. Choices B, C, and D incorrectly suggest the region is unbounded in various directions.

Question 8

What is the complete set of corner points for the unbounded feasible region defined by the system of inequalities $x \ge 0$ , $y \ge 0$ , $2x + y \ge 8$ , and $x + 3y \ge 9$ ?

{(0,8), (3,2), (9,0)} (correct answer)
{(0,0), (0,8), (3,2), (9,0)}
{(0,3), (4,0), (3,2)}
{(0,8), (9,0)}

Explanation: The feasible region is in the first quadrant (

x \ge 0, y \ge 0

) and lies above both lines

2x+y=8

and

x+3y=9

. The corner points are the vertices that form the lower boundary of this unbounded region. One corner point is the y-intercept of

2x+y=8

, which is (0,8), as this satisfies

0+3(8) \ge 9

. Another corner point is the x-intercept of

x+3y=9

, which is (9,0), as this satisfies

2(9)+0 \ge 8

. The third corner point is the intersection of the two lines. Solving the system

2x+y=8

and

x+3y=9

: From the first equation,

y=8-2x

. Substituting into the second gives

x+3(8-2x)=9 \implies x+24-6x=9 \implies -5x = -15 \implies x=3

. Then

y=8-2(3)=2

. The intersection is (3,2). The set of corner points is therefore {(0,8), (3,2), (9,0)}.

Question 9

For the feasible region defined by the system of inequalities $x \ge 0$ , $y \ge 0$ , $x+2y \le 12$ , and $3x+2y \le 24$ , what are the coordinates of the corner point with the largest x-coordinate?

(0, 6)
(6, 3)
(8, 0) (correct answer)
(12, 0)

Explanation: First, we must identify all corner points of the feasible region. The vertices are (0,0), the x-intercepts, the y-intercepts, and the intersection of the boundary lines. The y-intercept is determined by the most restrictive constraint at

x=0

. From

x+2y \le 12

, we get

y \le 6

. From

3x+2y \le 24

, we get

y \le 12

. The more restrictive is

y \le 6

, so (0,6) is a vertex. The x-intercept is determined by the most restrictive constraint at

y=0

. From

x+2y \le 12

, we get

x \le 12

. From

3x+2y \le 24

, we get

x \le 8

. The more restrictive is

x \le 8

, so (8,0) is a vertex. The intersection of

x+2y=12

and

3x+2y=24

is found by subtracting the first equation from the second:

2x=12 \implies x=6

. Then

6+2y=12 \implies 2y=6 \implies y=3

. So (6,3) is a vertex. The corner points are (0,0), (0,6), (8,0), and (6,3). Comparing their x-coordinates (0, 0, 8, 6), the largest is 8. The corner point with the largest x-coordinate is (8,0).

Question 10

Which of the following points is a vertex of the feasible region for the system: $y \ge 2$ , $x \le 8$ , and $3x - 2y \ge 6$ ? Non-negativity constraints $x \ge 0$ and $y \ge 0$ also apply.

(2, 0)
(8, 2)
(10/3, 2) (correct answer)
(8, 9)

Explanation: The feasible region is defined by

x \ge 0

y \ge 2

x \le 8

, and

3x-2y \ge 6

. We find the corner points by finding the intersections of the boundary lines and checking if they satisfy all inequalities.

Intersection of $y=2$ and $3x-2y=6$ : Substitute $y=2$ into the equation: $3x-2(2)=6 \implies 3x-4=6 \implies 3x=10 \implies x=10/3$ . The point is (10/3, 2). Check constraints: $10/3 \le 8$ (true), $2 \ge 2$ (true), $10/3 \ge 0$ (true). This is a vertex.
Intersection of $x=8$ and $3x-2y=6$ : Substitute $x=8$ : $3(8)-2y=6 \implies 24-2y=6 \implies 18=2y \implies y=9$ . The point is (8, 9). Check constraints: $9 \ge 2$ (true), $8 \le 8$ (true). This is also a vertex.
Intersection of $x=8$ and $y=2$ : The point is (8,2). Check constraint $3x-2y \ge 6$ : $3(8)-2(2) = 24-4 = 20 \ge 6$ (true). This is also a vertex. Of the choices given, only (10/3, 2) is listed.

Question 11

How many corner points (vertices) does the feasible region defined by the system $x \ge 0$ , $y \ge 0$ , $x \le 2$ , $3x+y \le 9$ , and $x+2y \le 10$ have?

3
4
5 (correct answer)
6

Explanation: The corner points are the intersections of the boundary lines that satisfy all inequalities. The boundary lines are

x=0, y=0, x=2, 3x+y=9, x+2y=10

. We find the vertices systematically:

(0,0) is a vertex.
On the y-axis ( $x=0$ ): The constraints become $y \ge 0$ , $y \le 9$ , and $2y \le 10$ (or $y \le 5$ ). The most restrictive is $y \le 5$ . So, (0,5) is a vertex.
On the x-axis ( $y=0$ ): The constraints become $x \ge 0$ , $x \le 2$ , $3x \le 9$ (or $x \le 3$ ), and $x \le 10$ . The most restrictive is $x \le 2$ . So, (2,0) is a vertex.
On the line $x=2$ : The constraints on y are $y \ge 0$ , $6+y \le 9$ (or $y \le 3$ ), and $2+2y \le 10$ (or $y \le 4$ ). The most restrictive is $y \le 3$ . So, (2,3) is a vertex.
Intersection of non-axis/non-constant boundary lines: $3x+y=9$ and $x+2y=10$ . Solving gives $y=9-3x$ , so $x+2(9-3x)=10 \implies x+18-6x=10 \implies -5x=-8 \implies x=8/5$ . Then $y=9-3(8/5) = 21/5$ . The point (8/5, 21/5) satisfies all other constraints ( $0 \le 8/5 \le 2$ , $0 \le 21/5$ ). So, (8/5, 21/5) is a vertex. The vertices are (0,0), (0,5), (2,0), (2,3), and (8/5, 21/5). There are a total of 5 corner points.

Question 12

A feasible region is defined by the inequalities $x \ge 0$ , $y \ge 0$ , $x + 2y \le 10$ , $3x + y \le 15$ , and $x + y \le 8$ . Which of the following inequalities is redundant for defining this region?

$x \ge 0$
$y \ge 0$
$3x + y \le 15$
$x + y \le 8$ (correct answer)

Explanation: To determine if a constraint is redundant, we find the feasible region defined by the other constraints and check if the potential redundant constraint is satisfied. The corner points of the region defined by

x \ge 0

y \ge 0

x + 2y \le 10

, and

3x + y \le 15

are (0,0), (5,0), (0,5), and the intersection of

x+2y=10

and

3x+y=15

. Solving this system gives

y=15-3x

, so

x+2(15-3x)=10 \implies x+30-6x=10 \implies -5x=-20 \implies x=4

. Then

y=15-3(4)=3

. The intersection point is (4,3). Now we check if the constraint

x+y \le 8

is satisfied by all these corner points: (0,0):

0+0=0 \le 8

. (5,0):

5+0=5 \le 8

. (0,5):

0+5=5 \le 8

. (4,3):

4+3=7 \le 8

. Since all corner points of the feasible region satisfy this inequality, the entire region is contained within the half-plane

x+y \le 8

, making this constraint redundant.