Linear Inequalities & Feasible Regions - Finite Mathematics

A dietitian is planning meals with constraints: $3p + 2c \geq 24$ (protein requirement), $p + 4c \geq 20$ (calorie requirement), where $p$ represents protein servings and $c$ represents carbohydrate servings. If the dietitian wants to minimize $p + c$ while staying within the feasible region, which corner point should be evaluated first?
Answer: The intersection of $3p + 2c = 24$ and $p + 4c = 20$ For linear programming, the optimal solution occurs at a corner point of the feasible region. Since we're minimizing $p + c$ , we need the corner point that's closest to the origin while satisfying both $\geq$ constraints. The intersection of the two constraint lines $3p + 2c = 24$ and $p + 4c = 20$ gives the corner point where both constraints are tight. Choice B and C give boundary points but not necessarily optimal. Choice D is incorrect because $(0,0)$ violates both constraints since we need $\geq$ relationships.
A feasible region is defined by $x \ge 0$ , $y \ge 0$ , and two other linear inequalities. Two of the corner points of the region are $(0,6)$ and $(8,0)$ . The other two corner points are $(0,0)$ and $(4,4)$ . Which of the following systems defines this region?
Answer: $\begin{cases} x+y \le 8 \\ x+2y \le 12 \end{cases}$ The feasible region is a quadrilateral in the first quadrant with vertices $(0,0), (8,0), (4,4), (0,6)$ . We need to find the two lines that form the upper boundary. One line passes through $(8,0)$ and $(4,4)$ . Its slope is $(4-0)/(4-8) = -1$ . The equation is $y-0 = -1(x-8)$ , which is $x+y=8$ . The region is below this line, so $x+y \le 8$ . The other line passes through $(4,4)$ and $(0,6)$ . Its slope is $(6-4)/(0-4) = -1/2$ . The equation is $y-6 = -1/2(x-0)$ , which is $2y-12=-x$ , or $x+2y=12$ . The region is also below this line, so $x+2y \le 12$ . The correct system, including non-negativity, is $x \ge 0, y \ge 0, x+y \le 8, x+2y \le 12$ . Choice C provides the two non-trivial inequalities.
The feasible region of a system of linear inequalities consists of all points $(x,y)$ such that $x \ge 0$ , $y \ge 0$ , and $y \le mx+4$ , where $m$ is a constant. The area of the feasible region is 16 square units. What is the value of $m$ ?
Answer: $m = -1/2$ The region is bounded by $x=0$ (y-axis), $y=0$ (x-axis), and the line $y = mx+4$ . For the region to be bounded and form a finite area in the first quadrant, the line must have a negative slope, so $m<0$ . The y-intercept of the line is $(0,4)$ . The x-intercept is found by setting $y=0$ , which gives $0 = mx+4$ , so $x = -4/m$ . Since $m<0$ , the x-intercept is positive. The feasible region is a triangle with vertices at $(0,0)$ , $(0,4)$ , and $(-4/m, 0)$ . The height of the triangle is 4, and the base is $-4/m$ . The area is given by $A = \frac{1}{2} \times \text{base} \times \text{height}$ . We are given that the area is 16. So, $16 = \frac{1}{2} \times (-\frac{4}{m}) \times 4$ . This simplifies to $16 = -\frac{8}{m}$ . Solving for $m$ , we get $16m = -8$ , so $m = -8/16 = -1/2$ .
Which production plan $(x, y)$ is infeasible because it violates the finishing constraint, but satisfies all other constraints?
Answer: The plan $(20, 26)$ , which violates the finishing constraint only. We must test the point $(20, 26)$ against all constraints. Contract: $y=26 \ge 10$ (Satisfied). Assembly: $2(20) + 3(26) = 40 + 78 = 118 \le 120$ (Satisfied). Finishing: $x + 2y = 20 + 2(26) = 20 + 52 = 72$ . This is not $\le 70$ , so the finishing constraint is violated. Thus, $(20, 26)$ is infeasible due to the finishing constraint only. Distractor A violates only the assembly constraint. Distractor B violates two constraints. Distractor C is a feasible plan.
Which statement accurately describes the feasible region defined by the system $x \ge 3$ , $y \ge 2$ , and $x + y \ge 8$ ?
Answer: It is an unbounded region with two corner points. The region is defined by being to the right of the vertical line $x=3$ , above the horizontal line $y=2$ , and above the line $x+y=8$ . This region is unbounded upwards and to the right. The corner points are the vertices of the boundary of the feasible region. The intersection of $x=3$ and $y=2$ is $(3,2)$ . This point does not satisfy $x+y \ge 8$ because $3+2=5$ , so it is not in the region. The intersection of $x=3$ and $x+y=8$ is $(3,5)$ . This point satisfies $y \ge 2$ , so it is a corner point. The intersection of $y=2$ and $x+y=8$ is $(6,2)$ . This point satisfies $x \ge 3$ , so it is also a corner point. Therefore, the region is unbounded and has exactly two corner points: $(3,5)$ and $(6,2)$ .
The feasible region for a system of linear inequalities is a triangle with vertices at $(0, 0)$ , $(6, 0)$ , and $(4, 5)$ . Which of the following systems of inequalities defines this region?
Answer: $\begin{cases} y \ge 0 \\ 5x - 4y \ge 0 \\ 5x + 2y \le 30 \end{cases}$ The correct system of inequalities must define a triangular region with the given vertices. The boundary $y=0$ (the x-axis) contains the side from $(0,0)$ to $(6,0)$ , and the region is shaded above it, so $y \ge 0$ . The line passing through $(0,0)$ and $(4,5)$ has the equation $y = \frac{5}{4}x$ , which can be written as $5x - 4y = 0$ . Testing a point inside the region, like $(5,0)$ , gives $5(5) - 4(0) = 25 \ge 0$ , so the inequality is $5x - 4y \ge 0$ . The line passing through $(4,5)$ and $(6,0)$ has a slope of $\frac{0-5}{6-4} = -\frac{5}{2}$ . Its equation is $y - 0 = -\frac{5}{2}(x - 6)$ , which simplifies to $2y = -5x + 30$ , or $5x + 2y = 30$ . Testing a point inside the region, like $(4,0)$ , gives $5(4) + 2(0) = 20 \le 30$ , so the inequality is $5x + 2y \le 30$ . Combining these gives the system in choice A.
Consider the inequality $2x - 3y > 6$ . If this inequality is graphed on a coordinate plane, which of the following statements about the boundary line and shaded region is correct?
Answer: The boundary line has slope $\frac{2}{3}$ and the region below the line is shaded Rewriting $2x - 3y > 6$ in slope-intercept form: $-3y > -2x + 6$ , so $y < \frac{2}{3}x - 2$ . The boundary line $y = \frac{2}{3}x - 2$ has slope $\frac{2}{3}$ . Since the inequality is $y < \frac{2}{3}x - 2$ , we shade below the line. Choice A has the correct slope but wrong region. Choice C has the wrong slope ( $\frac{3}{2}$ instead of $\frac{2}{3}$ ). Choice D has the wrong slope ( $-\frac{2}{3}$ ) and wrong region.
The feasible region for a linear programming problem is defined by the system: $x + 2y \leq 10$ , $3x + y \leq 12$ , $x \geq 0$ , $y \geq 0$ . If a new constraint $x + y \geq 3$ is added to the system, what happens to the area of the feasible region?
Answer: The area decreases, and the new region is bounded The original feasible region includes the origin and forms a polygon. Adding $x + y \geq 3$ eliminates the portion near the origin where $x + y < 3$ , reducing the total area. The new region remains bounded because the original region was bounded. Choice A is incorrect because the reduction isn't exactly 3 square units. Choice C is wrong because $x + y \geq 3$ is not redundant—it eliminates feasible points. Choice D misunderstands that adding constraints typically reduces (not increases) the feasible region.
A linear programming problem has the feasible region bounded by $x + 3y \leq 15$ , $2x + y \leq 10$ , $x \geq 0$ , and $y \geq 0$ . If the objective function is $z = ax + by$ where $a > 0$ and $b > 0$ , under what condition will the optimal solution occur at the point $(3, 4)$ ?
Answer: When $\frac{a}{b} = \frac{1}{3}$ , making the objective function parallel to $x + 3y = 15$ In linear programming, when you need to find conditions for an optimal solution at a specific vertex, you're looking for when the objective function's slope makes that vertex optimal. The key insight is that a vertex is optimal when the objective function is parallel to one of the constraint lines forming that vertex. First, let's verify that (3, 4) is indeed a vertex of the feasible region. Checking the constraints: $3 + 3(4) = 15$ and $2(3) + 4 = 10$ , so this point lies at the intersection of both constraint lines $x + 3y = 15$ and $2x + y = 10$ . For the objective function $z = ax + by$ to have its optimum at (3, 4), the function must be parallel to one of these constraint lines. The objective function can be rewritten as $y = -\frac{a}{b}x + \frac{z}{b}$ , giving it a slope of $-\frac{a}{b}$ . The constraint $x + 3y = 15$ has slope $-\frac{1}{3}$ , while $2x + y = 10$ has slope $-2$ . For the objective function to be parallel to $x + 3y = 15$ , we need $-\frac{a}{b} = -\frac{1}{3}$ , which means $\frac{a}{b} = \frac{1}{3}$ . Answer A ( $a = b$ ) gives slope $-1$ , which isn't parallel to either constraint. Answer B ( $\frac{a}{b} = 2$ ) makes the objective function parallel to $2x + y = 10$ , but this would optimize at a different vertex. Answer D ( $\frac{a}{b} = \frac{3}{4}$ ) gives slope $-\frac{3}{4}$ , which matches neither constraint line. Remember: optimal solutions occur at vertices where the objective function is parallel to a constraint boundary.
Which of the following points lies within the feasible region defined by the system $x > 2$ , $y \ge 1$ , and $x + y < 8$ ?
Answer: The point $(3, 4)$ , which satisfies all three inequalities. We must check which point satisfies all three inequalities simultaneously. For point $(3, 4)$ : $3 > 2$ (True), $4 \ge 1$ (True), and $3 + 4 = 7 < 8$ (True). So, $(3,4)$ is in the feasible region. For point $(2,5)$ , $x=2$ fails the strict inequality $x>2$ . For point $(4,4)$ , $x+y=8$ fails the strict inequality $x+y<8$ . For point $(7,1)$ , $x+y=8$ fails the strict inequality $x+y<8$ . The distractors are chosen to be on the boundaries defined by strict inequalities, which are common points of confusion.
Consider the system of inequalities: $-x + 2y \leq 4$ , $x + y \geq 3$ , $x \leq 5$ , $y \geq 0$ . Which of the following points lies on the boundary of the feasible region but is NOT a corner point?
Answer: $(1, 2)$ on the line $x + y = 3$ When working with systems of inequalities, you need to distinguish between boundary points and corner points of the feasible region. A boundary point lies on the edge of any constraint line, while a corner point is where exactly two boundary lines intersect to form a vertex of the feasible region. Let's verify that option D, $(1, 2)$ , satisfies all constraints: $-1 + 2(2) = 3 \leq 4$ ✓, $1 + 2 = 3 \geq 3$ ✓, $1 \leq 5$ ✓, and $2 \geq 0$ ✓. This point lies exactly on the line $x + y = 3$ (since $1 + 2 = 3$ ), making it a boundary point. However, it's not a corner point because it doesn't represent the intersection of two constraint boundaries that form a vertex of the feasible region. Now let's examine why the other options are corner points. Option A: $(2, 3)$ lies at the intersection of $-x + 2y = 4$ and $x + y = 3$ , making it a corner point. Option B: $(3, 0)$ is where $y = 0$ meets $x + y = 3$ , also a corner point. Option C: $(5, 1)$ sits at the intersection of $x = 5$ and another boundary line, making it a corner point as well. The key insight is that corner points occur where exactly two constraint boundaries intersect to create vertices, while other boundary points simply lie along the edges between corners. When identifying feasible regions, always check whether a boundary point represents a vertex intersection or just falls along an edge.
What is the area of the feasible region defined by the system of inequalities $x \ge 1$ , $y \ge 2$ , and $x + y \le 7$ ?
Answer: 8 square units The feasible region is a triangle bounded by the lines $x=1$ , $y=2$ , and $x+y=7$ . To find the area, we first need to find the vertices of this triangle. The vertices are the intersection points of these lines. Intersection of $x=1$ and $y=2$ is $(1,2)$ . Intersection of $x=1$ and $x+y=7$ is $(1,6)$ . Intersection of $y=2$ and $x+y=7$ is $(5,2)$ . The triangle is a right triangle with vertices $(1,2)$ , $(1,6)$ , and $(5,2)$ . The length of the vertical leg is the difference in y-coordinates: $6-2=4$ . The length of the horizontal leg is the difference in x-coordinates: $5-1=4$ . The area of a right triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8$ square units.
A farmer has constraints $x + y \leq 100$ (land), $2x + y \leq 150$ (labor), and $x, y \geq 0$ , where $x$ and $y$ represent acres of two crops. If an additional constraint $y \geq 20$ is imposed due to crop rotation requirements, what is the new corner point that was not in the original feasible region?
Answer: $(0, 20)$ representing minimum crop rotation requirement When you encounter linear programming problems with changing constraints, focus on how new restrictions modify the feasible region's boundaries and create new corner points. To find the new corner point, you need to identify where the additional constraint $y \geq 20$ intersects with the existing boundaries. The original feasible region has corner points at $(0,0)$ , $(0,100)$ , $(50,50)$ , and $(75,0)$ formed by intersections of $x + y = 100$ , $2x + y = 150$ , and the axes. The new constraint $y = 20$ creates a horizontal line that cuts through the original region. This line intersects the y-axis at $(0,20)$ , which becomes a new corner point. You can verify this satisfies all constraints: $0 + 20 = 20 \leq 100$ ✓, $2(0) + 20 = 20 \leq 150$ ✓, and $y = 20 \geq 20$ ✓. Option A $(30,20)$ is incorrect because this point lies on the line $y = 20$ but isn't a corner point—it's in the interior of an edge, not at the intersection of two constraint boundaries. Option B $(50,50)$ is wrong because this was already a corner point in the original feasible region before adding $y \geq 20$ . Option D $(130,20)$ violates the land constraint since $130 + 20 = 150 > 100$ . Remember: when constraints are added to linear programming problems, new corner points occur where the new constraint boundary intersects existing boundaries of the feasible region, typically at edges or vertices.
Let $R$ be the feasible region for the system $x \ge 0$ , $y \ge 0$ , and $x + y \le 10$ . A new constraint, $2x + y \ge 8$ , is added to form a new region, $R'$ . How does the new feasible region, $R'$ , compare to the original region $R$ ?
Answer: $R'$ is a quadrilateral formed by removing a triangular section from $R$ . The original region $R$ is a triangle with vertices $(0,0)$ , $(10,0)$ , and $(0,10)$ . The new constraint $2x+y \ge 8$ corresponds to the half-plane above the line $2x+y=8$ . This line passes through $(4,0)$ and $(0,8)$ . Since both these points are on the boundary of the original region $R$ , the new constraint is not redundant. It removes the portion of $R$ that is below the line $2x+y=8$ , which is a small triangle with vertices $(0,0)$ , $(4,0)$ , and $(0,8)$ . The remaining feasible region $R'$ is a quadrilateral with vertices $(4,0)$ , $(10,0)$ , $(0,10)$ , and $(0,8)$ .
For which of the following values of a positive constant $k$ does the additional constraint $x \ge k$ result in a feasible region with exactly three corner points?
Answer: $k=2$ , which results in a triangular region. First, find the vertices of the original region. They are $(0,0)$ , $(3,0)$ (from $3x+y=9$ ), $(0,4)$ (from $x+2y=8$ ), and the intersection of $3x+y=9$ and $x+2y=8$ . Solving this system gives $y=9-3x$ , so $x+2(9-3x)=8 \Rightarrow x+18-6x=8 \Rightarrow 5x=10 \Rightarrow x=2$ . Then $y=3$ . The vertex is $(2,3)$ . The original region is a quadrilateral with vertices $(0,0)$ , $(3,0)$ , $(2,3)$ , $(0,4)$ . The constraint $x \ge k$ is a vertical line. If $k=2$ , this line passes through the vertex $(2,3)$ . The new vertices will be the intersection of $x=2$ with the x-axis, which is $(2,0)$ ; the vertex $(3,0)$ which satisfies $x \ge 2$ ; and the vertex $(2,3)$ itself. These three points, $(2,0)$ , $(3,0)$ , and $(2,3)$ , form a triangle. Thus, for $k=2$ , the feasible region has exactly three corner points.
A company's production is constrained by $x + 2y \leq 12$ and $2x + y \leq 10$ , where $x, y \geq 0$ . If the constraint $x + 2y \leq 12$ is changed to $x + 2y \leq 8$ , which corner point of the original feasible region becomes infeasible?
Answer: $(0, 6)$ First, find the corner points of the original region. The vertices are: $(0,0)$ , $(0,6)$ from $x + 2y = 12$ , $(5,0)$ from $2x + y = 10$ , and $(4,2)$ from the intersection of both constraints. When the constraint changes to $x + 2y \leq 8$ , check each point: $(0,6)$ gives $0 + 2(6) = 12 > 8$ , so it's infeasible. Points $(4,2)$ , $(5,0)$ , and $(0,0)$ all satisfy $x + 2y \leq 8$ . The point $(0,6)$ was on the original boundary but violates the new tighter constraint.
Consider the feasible region defined by the system of inequalities: $x \ge 0$ , $y \ge 0$ , $x \le 5$ , $y \le 5$ , and $x + y \le 12$ . Which of the following inequalities is redundant, meaning its removal does not alter the feasible region?
Answer: $x + y \le 12$ , because the other constraints already restrict the region sufficiently. The constraints $x \ge 0$ , $y \ge 0$ , $x \le 5$ , and $y \le 5$ define a square region with vertices at $(0,0)$ , $(5,0)$ , $(5,5)$ , and $(0,5)$ . To determine if $x + y \le 12$ is redundant, we can check if the entire square satisfies this condition. The maximum value of $x+y$ within this square occurs at the vertex $(5,5)$ . At this point, $x+y = 5+5=10$ . Since $10 \le 12$ , all points within the square satisfy the inequality $x+y \le 12$ . Therefore, this inequality is redundant and its removal does not change the feasible region.

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