For the constraint 3x + 2y ≤ 18, the point (4,4) is
A Feasible for it
B Infeasible for it
C On boundary
D Always optimal
Substitute x = 4, y = 4: 3(4) + 2(4) = 12 + 8 = 20, which is not ≤ 18. Wait—so it violates the constraint, hence infeasible for this constraint.
When checking feasibility of a point, you must test it in
A Only objective
B Only two constraints
C Every constraint
D Only nonnegativity
A point is feasible only if it satisfies all constraints and non-negativity. Even if it satisfies most constraints, failing one inequality makes it infeasible and it cannot be an LPP solution.
In Z = 7x + 5y, if x increases by 1 and y fixed, Z changes by
A +5
B +12
C +2
D +7
The coefficient of x tells the change in Z per unit change in x when other variables stay constant. Here, increasing x by 1 increases Z by 7 units.
For Z = 6x + 3y, slope of objective line is
A -1/2
B 2
C -2
D 1/2
Write 6x + 3y = Z ⇒ y = -(6/3)x + Z/3 = -2x + Z/3. So slope is -2, determined by the ratio of coefficients.
If feasible region is bounded and nonempty, then a maximum value
A Never exists
B Must exist
C Only sometimes
D Is infinite
For a linear objective function over a nonempty bounded closed feasible region, the maximum and minimum are attained at some feasible point, typically at a vertex.
A point (x,y) lies on boundary of 2x + y ≤ 10 when
A 2x + y = 10
B 2x + y < 10
C 2x + y > 10
D 2x + y ≠ 10
The boundary line of a ≤ constraint is formed by replacing ≤ with equality. Points on this line satisfy the constraint tightly and may act as edges of the feasible region.
For constraints x + y ≤ 6 and x + y ≤ 8, the second is
A Inconsistent
B Binding always
C Redundant
D Unbounded
If x + y ≤ 6 is enforced, then x + y ≤ 8 automatically holds. So the second constraint adds no restriction to feasible region, making it redundant.
If constraints are x ≥ 0, y ≥ 0, and x + y ≤ 0, feasible region is
A Empty set
B Entire quadrant
C A line segment
D Only origin
In first quadrant, x and y are nonnegative. The condition x + y ≤ 0 forces x = 0 and y = 0 only. So the feasible region reduces to a single point.
A feasible region that is a single point implies the LPP has
A Unique feasible solution
B Unbounded solution
C Alternate optimum
D No feasible solution
If feasible set contains exactly one point, then there is exactly one feasible solution. That point becomes automatically optimal for both maximization and minimization since no alternatives exist.
If two boundary lines intersect outside the feasible region, that intersection is
A Feasible vertex
B Infeasible vertex
C Optimal vertex
D Binding point
Intersection of boundary lines gives a candidate corner point, but it must satisfy all constraints. If it violates any inequality, it lies outside feasible region and is infeasible.
In graphical LPP, the “optimal vertex” is chosen by
A Smallest x value
B Largest y value
C Best objective value
D Nearest origin
After finding feasible vertices, compute the objective value at each. The vertex giving maximum (or minimum) Z is optimal. Distance from origin alone is not a valid rule.
For inequality x − y ≥ 2, converting to ≤ form gives
A -x + y ≤ -2
B x + y ≤ 2
C x − y ≤ 2
D -x − y ≤ -2
Multiply both sides by -1 to reverse the inequality: -(x − y) ≤ -2 ⇒ -x + y ≤ -2. This helps standardization where many constraints are written as ≤.
The inequality y ≥ 0 ensures the feasible points are
A On/below x-axis
B On/above x-axis
C On y-axis only
D In third quadrant
y ≥ 0 includes all points with nonnegative y-coordinate, i.e., on or above the x-axis. Combined with x ≥ 0, it restricts to the first quadrant region.
For constraint y ≤ 5, feasible points are
A On/above y=5
B Left of y=5
C On/below y=5
D Right of y=5
The boundary is the horizontal line y = 5. The inequality y ≤ 5 includes the line and all points with y-coordinate less than or equal to 5.
If objective function is parallel to one constraint edge, it may create
A Infeasible region
B Redundant variables
C No vertices
D Alternate optima
When objective line coincides with a boundary segment at optimum, every point on that segment gives same Z. This creates multiple optimal solutions instead of a single optimal point.
For Z = 2x + 3y, iso-profit lines are
A Parallel lines
B Intersecting circles
C Curved arcs
D Parabola family
Setting Z = k gives 2x + 3y = k, which is a straight line. Different k values shift the line parallel. This family is used to locate optimum.
If Z = 4x + 0y, iso-profit lines are
A Horizontal lines
B Slant lines
C Vertical lines
D Curved lines
Z = 4x implies x = Z/4. For fixed Z, x is constant, giving vertical lines. So objective depends only on x, and optimum occurs at extreme feasible x-values.
If Z = 0x + 5y, iso-profit lines are
A Vertical lines
B Slant lines
C Horizontal lines
D Circle arcs
Z = 5y gives y = Z/5. For constant Z, y is constant, producing horizontal lines. Optimum depends on extreme feasible y-values within the feasible region.
In corner point evaluation, if a vertex gives same Z as another vertex, it suggests
A Unbounded region sure
B Alternative optimum possible
C Infeasible constraints
D Redundant objective
Equal best values at two distinct feasible vertices often indicate an entire edge between them may also be optimal, provided that edge lies on a boundary where objective is parallel.
If feasible region is empty, then the LPP has
A Infinite solutions
B Unique optimum
C Alternate optimum
D No feasible solution
Empty feasible region means no point satisfies all constraints together. Without any feasible solution, the objective cannot be optimized, so the LPP is infeasible.
In LPP, the “standard form” (intro) usually includes
A Equality constraints
B Quadratic constraints
C Negative variables
D Trig objectives
In introductory standard form for simplex, inequalities are converted into equalities using slack or surplus variables. This creates a system of linear equations suitable for tableau methods.
Adding slack variables keeps feasibility because slack must be
A Negative only
B Complex number
C Nonnegative
D Always zero
Slack represents unused resources, so it cannot be negative. Writing slack as a nonnegative variable preserves the meaning of ≤ constraints and ensures the converted equality matches original inequality.
A redundant constraint can still be “binding” at optimum?
A No, never
B Yes, possible
C Only in 3D
D Only in infeasible
A constraint may be redundant overall (does not change feasible region) but still hold as equality at the optimum point. Even then, removing it does not change feasibility set.
If objective is minimized and feasible region is bounded, then minimum value
A Never exists
B Is always zero
C Is infinite
D Must exist
For a linear objective over a nonempty bounded feasible region, both maximum and minimum exist. The minimum occurs at at least one vertex, by the corner point theorem.
If constraints allow x and y to increase without limit, feasible region is
A Bounded
B Empty
C Unbounded
D Single point
Unbounded feasible region means there are feasible points arbitrarily far away. Whether optimum is unbounded depends on objective direction. Missing constraints often lead to unbounded feasible sets.
In a maximization LPP, if objective increases in an unbounded feasible direction, then maximum is
A Unbounded
B Unique
C At origin
D At midpoint
If there exists a feasible ray along which Z keeps increasing, there is no finite maximum. This is an unbounded optimum situation and indicates constraints do not limit improvement.
Which point is feasible for x + y ≤ 5, x ≥ 0, y ≥ 0?
A (4,3)
B (5,2)
C (3,4)
D (2,2)
Check sums: (2,2) gives 4 ≤ 5, so feasible. (4,3) gives 7, (5,2) gives 7, (3,4) gives 7, all violate x + y ≤ 5.
For constraint 2x + y ≤ 8, which point lies on boundary?
A (2,3)
B (3,2)
C (1,1)
D (4,1)
Substitute: 2(3)+2 = 8, so it lies exactly on boundary. Others give 7, 3, and 9 respectively, so (2,3) and (1,1) are inside, (4,1) outside.
If (x,y) = (0,6) is tested for x + y ≤ 5, it is
A Feasible
B Optimal always
C Infeasible
D Binding always
Substitute: 0 + 6 = 6 ≤ 5 is false, so (0,6) violates the constraint. It cannot be in feasible region even though it satisfies x ≥ 0 and y ≥ 0.
The constraint x ≤ 4 represents the region
A Left of x=4
B Right of x=4
C Above x=4
D Below x=4
Boundary is vertical line x = 4. The inequality x ≤ 4 includes all points with x-coordinate less than or equal to 4, i.e., the half-plane to the left.
In LPP formulation, “units” matter because constraints must be
A Randomly adjusted
B Always unitless
C Dimensionally consistent
D Always integers
When writing constraints from word problems, both sides must represent the same unit (hours, kg, rupees, etc.). Wrong units cause incorrect coefficients and wrong feasible region.
If a constraint is written incorrectly, the most direct result is
A Same optimum always
B Wrong feasible region
C Better solution always
D More vertices always
Constraints define the feasible set. Any error in inequality sign or coefficients changes the feasible region, causing wrong vertices and wrong optimal solution, even if objective function is correct.
A feasible region edge corresponds to
A A binding constraint
B A slack variable
C An objective line
D A random line
Each boundary edge of feasible polygon lies on a constraint line where the inequality is satisfied as equality. Along that edge, the corresponding constraint is binding.
When selecting optimal point by moving line, you stop at
A First axis touch
B Random vertex
C Last feasible touch
D Any interior point
For maximization, keep shifting iso-profit line to increase Z. The last position that still intersects the feasible region gives optimal value; further shift would leave feasible region.
A system of constraints that intersect in a strip-like region is typically
A Single point
B Empty region
C Circular region
D Unbounded region
A strip extends indefinitely in some direction, so the feasible region is unbounded. Such regions can still have optimal solutions if objective is limited by direction, but not always.
In 2-variable LPP, if feasible region is a triangle, number of vertices is
A 2
B 3
C 4
D 5
A triangle has three corner points. In corner-point method, you evaluate objective at these three feasible vertices and choose the maximum or minimum depending on the problem.
If objective Z = 3x + 2y is evaluated at (2,1), Z equals
A 7
B 5
C 8
D 6
Substitute x = 2, y = 1: Z = 3(2) + 2(1) = 6 + 2 = 8. So correct value is 8, meaning option A should be correct.
A “feasible corner point” must satisfy
A All constraints
B Only two constraints
C Only objective
D Only axes
Corner points are found by intersection of boundary lines, but only those that satisfy every constraint and non-negativity are feasible vertices. Only feasible vertices are used for optimization.
If constraints include x ≥ 0, y ≥ 0 and y ≥ 2, feasible region starts
A Below y=2
B Left of y=2
C Right of y=2
D Above y=2
y ≥ 2 restricts the region on or above the horizontal line y = 2. With x ≥ 0 and y ≥ 0, feasible points lie in first quadrant but at least 2 units above x-axis.
If objective line is perpendicular to x-axis, it means objective depends only on
A y only
B x and y
C x only
D neither
A line perpendicular to x-axis is vertical, meaning x is constant. Objective lines vertical correspond to Z depending only on x, not y. So the dependence stated must match.
The “slope of objective line” helps to
A Count constraints
B Compare with edges
C Find intercepts only
D Remove variables
Comparing objective line slope with boundary edges helps detect alternate optimum (parallel edge) or identify direction of improvement. It guides how iso-profit/iso-cost lines move to touch feasible region.
A common condition for multiple optimal solutions is
A Feasible region empty
B Region unbounded only
C Objective parallel to edge
D Objective nonlinear
Multiple optima occur when objective line at optimal value overlaps a feasible boundary segment. This happens when the objective slope matches the slope of a boundary edge.
If a feasible region is unbounded but objective is minimized, minimum may still
A Exist
B Never exist
C Be infinite only
D Be zero always
Even with unbounded feasible region, a minimum can exist if objective decreases in a direction that is blocked by constraints. Unbounded region alone does not guarantee unbounded optimum.
The “graphical limitation” of LPP mainly refers to
A Using straight lines
B Using inequalities
C Using vertices
D More than two variables
Graphical method works smoothly for two variables. With three variables it needs 3D graphs, and for more variables visualization becomes impractical, so simplex or computational methods are used.
A solution is called “optimal feasible” when it is
A Infeasible but best
B Feasible but random
C Feasible and best
D Infeasible and worst
Optimal feasible means it satisfies all constraints and gives the best objective value (maximum or minimum). Feasibility ensures it is allowed; optimality ensures it is the best among allowed points.
When two constraints intersect on the axes, that intersection is a
A Corner candidate
B Redundant point
C Impossible point
D Shadow point
Intersections on axes are still intersection points of boundary lines. If they satisfy all constraints, they become feasible vertices. Many feasible regions include axis intercept corners due to non-negativity.
The term “iso-cost line” is used mainly in
A Maximization problems
B Minimization problems
C Infeasible problems
D Redundant problems
Iso-cost lines represent constant cost values for a cost objective function. In minimization, these lines are shifted toward smaller values until they just touch the feasible region.
If constraint coefficients are scaled (multiply by 2), the feasible half-plane
A Flips side
B Becomes curve
C Stays same
D Disappears
Multiplying both sides of a linear inequality by a positive number does not change the set of solutions. The same half-plane is represented; only the equation’s numerical form changes.
A “production problem” LPP often includes constraints on
A Resources available
B Random demand
C Circle radius
D Trig angles
Production LPP models limits like machine hours, labor time, raw materials, and budget. These become linear constraints. Objective typically maximizes profit or minimizes cost.
To detect unbounded optimum graphically, you look for objective lines
A Touching all vertices
B Rotating randomly
C Becoming circles
D Moving without limit
If you can shift the objective line in improving direction and it keeps intersecting feasible region without a final touch point, then Z can improve indefinitely, indicating unbounded optimum.