Linear Programming for CUET Maths

AnswerB · $2x + 3y \le 12$

1. Write each resource as (amount used per unit) × (number of units), added over the products.

2. 'At most' or 'available' gives ≤, 'at least' or 'must give' gives ≥; a profit is maximised and a cost minimised.

AnswerD · $5x + 2y \ge 40$

1. Write each resource as (amount used per unit) × (number of units), added over the products.

2. 'At most' or 'available' gives ≤, 'at least' or 'must give' gives ≥; a profit is maximised and a cost minimised.

AnswerA · $\text{maximise } Z = 50x + 30y$

1. Write each resource as (amount used per unit) × (number of units), added over the products.

2. 'At most' or 'available' gives ≤, 'at least' or 'must give' gives ≥; a profit is maximised and a cost minimised.

AnswerA · $x \ge y$

1. Write each resource as (amount used per unit) × (number of units), added over the products.

2. 'At most' or 'available' gives ≤, 'at least' or 'must give' gives ≥; a profit is maximised and a cost minimised.

AnswerB · Decision variables

1. $x$ and $y$ stand for the quantities to be decided, such as how many units to make.

2. They are the decision variables; $Z$ is the objective function and the inequalities are the constraints.

AnswerA · Both A and R are true, and R is the correct explanation of A.

1. $x$ and $y$ count units, which cannot be negative, so R is true.

2. That is exactly why the non-negative constraints are written: A is true and R explains it.

AnswerD · $x + y \le 10$

1. Write each resource as (amount used per unit) × (number of units), added over the products.

2. 'At most' or 'available' gives ≤, 'at least' or 'must give' gives ≥; a profit is maximised and a cost minimised.

AnswerB · $3x + y \le 15,\ x + 2y \le 10$

1. Write each resource as (amount used per unit) × (number of units), added over the products.

2. 'At most' or 'available' gives ≤, 'at least' or 'must give' gives ≥; a profit is maximised and a cost minimised.

AnswerD · $\text{minimise } Z = 400x + 700y$

1. Write each resource as (amount used per unit) × (number of units), added over the products.

2. 'At most' or 'available' gives ≤, 'at least' or 'must give' gives ≥; a profit is maximised and a cost minimised.

AnswerC · I, II and III only

1. I, II and III are the definitions: a linear objective, non-negative variables, and a feasible region where every constraint holds.

2. IV is false: a point of the feasible region is a feasible solution; an optimal solution is one that gives the best value of $Z$.

AnswerC · $x + y \le 60,\ x \ge 2y$

1. Write each resource as (amount used per unit) × (number of units), added over the products.

2. 'At most' or 'available' gives ≤, 'at least' or 'must give' gives ≥; a profit is maximised and a cost minimised.

AnswerA · $2x + 4y \ge 20,\ 3x + y \ge 12$

1. Write each resource as (amount used per unit) × (number of units), added over the products.

2. 'At most' or 'available' gives ≤, 'at least' or 'must give' gives ≥; a profit is maximised and a cost minimised.

AnswerB · Only II

1. I is false: the term $xy$ makes $Z$ non-linear; the objective must be linear.

2. II is true: it is a linear inequality in $x$ and $y$.

AnswerD · $\text{maximise } Z = 300x + 100y$

1. Write each resource as (amount used per unit) × (number of units), added over the products.

2. 'At most' or 'available' gives ≤, 'at least' or 'must give' gives ≥; a profit is maximised and a cost minimised.

AnswerB · $x + y \le 50,\ 4x + 6y \le 240$

1. Write each resource as (amount used per unit) × (number of units), added over the products.

2. 'At most' or 'available' gives ≤, 'at least' or 'must give' gives ≥; a profit is maximised and a cost minimised.

AnswerC · $x \ge 2y$

1. Write each resource as (amount used per unit) × (number of units), added over the products.

2. 'At most' or 'available' gives ≤, 'at least' or 'must give' gives ≥; a profit is maximised and a cost minimised.

AnswerB · $(2, 3)$

1. A point is feasible when it satisfies every constraint. Check $(2, 3)$: $x + y = 5$; $2x + y = 7$.

2. Each of the other points breaks at least one constraint.

AnswerC · The line $2x + 3y = 12$ and the side of it that contains the origin

1. A linear inequality describes a half-plane: the boundary line and everything on one side of it.

2. Test the origin: $2(0) + 3(0) = 0 \le 12$ is true, so the side with the origin is the one shaded.

AnswerB · $4$

1. Find where the boundary lines, the axes included, meet, and keep the meeting points that satisfy every constraint.

2. They are $(0, 0)$, $(0, 2)$, $(3, 1)$, $(4, 0)$: 4 in all.

AnswerD · $(3, 2)$

1. Put each point into every constraint.

2. $(3, 2)$: $3 + 2 = 5 \le 8$ and $3 + 6 = 9 \ge 6$. $(1, 1)$ and $(0, 1)$ fail $x + 3y \ge 6$; $(6, 4)$ fails $x + y \le 8$.

AnswerA · $(1, 3)$

1. At $(1, 3)$: $x - y = -2$, which is less than $-1$, so it breaks the second constraint.

2. Each of the other three satisfies all four constraints.

AnswerD · $(2, 2)$

1. A point is feasible when it satisfies every constraint. Check $(2, 2)$: $x + 2y = 6$; $x + y = 4$.

2. Each of the other points breaks at least one constraint.

AnswerA · $5$

1. Find where the boundary lines, the axes included, meet, and keep the meeting points that satisfy every constraint.

2. They are $(0, 2)$, $(0, 10)$, $(2, 0)$, $(6, 0)$, $(6, 4)$: 5 in all.

AnswerD · $x + y \le 5,\ x \le 3,\ x \ge 0,\ y \ge 0$

1. The region contains the origin, so both boundary lines are '$\le$' toward it.

2. Its corners $(0, 0)$, $(3, 0)$, $(3, 2)$, $(0, 5)$ lie on $x + y = 5$, $x = 3$ and the axes.

AnswerA · Yes, it is bounded

1. $x \le 4$ and $y \le 5$ keep every point inside the box $0 \le x \le 4$, $0 \le y \le 5$.

2. $x + y \ge 3$ only cuts off the corner near the origin, so the region is a bounded polygon.

AnswerC · The line $x - y = 2$ and the side of it away from the origin

1. Test the origin: $0 - 0 \ge 2$ is false, so the origin is not in the region.

2. The region is the half-plane on the far side of the line, and '$\ge$' includes the line itself.

AnswerC · $5$

1. The corners are $(0, 0)$, $(6, 0)$, $(6, 3)$, $(2, 5)$, $(0, 5)$.

2. The line $x + 2y = 12$ cuts off the corner $(6, 5)$ of the box, leaving 5 corners.

AnswerB · $(3, 3)$

1. The lines $x + y = 6$ and $y = x$ meet at $(3, 3)$, which satisfies every constraint.

2. $(0, 6)$ and $(2, 4)$ break $x \ge y$; $(6, 6)$ breaks $x + y \le 6$. The corners are $(0, 0)$, $(6, 0)$, $(3, 3)$.

AnswerD · $12$

1. The corner points of the feasible region are $(0, 0)$, $(0, 2)$, $(3, 1)$, $(4, 0)$.

2. Values of $Z$: $0$ at $(0, 0)$, $4$ at $(0, 2)$, $11$ at $(3, 1)$, $12$ at $(4, 0)$.

3. The region is bounded, so the maximum is the largest of these: $12$, at $(4, 0)$.

AnswerC · $24$

1. For a bounded region, the maximum is at a corner: work out $Z$ at each.

2. $Z = 0$, 20, $12 + 12 = 24$ and 18, so the maximum is 24, at $(3, 4)$.

AnswerA · $4$

1. Evaluate $Z$ at each corner: $4$ at $(0, 2)$, $20$ at $(4, 0)$, $36$ at $(6, 3)$, $22$ at $(2, 6)$.

2. The smallest is 4, at $(0, 2)$.

AnswerC · $26$

1. The corner points of the feasible region are $(0, 0)$, $(0, 5)$, $(2, 4)$, $(4, 0)$.

2. Values of $Z$: $0$ at $(0, 0)$, $20$ at $(0, 5)$, $26$ at $(2, 4)$, $20$ at $(4, 0)$.

3. The region is bounded, so the maximum is the largest of these: $26$, at $(2, 4)$.

AnswerB · $2$

1. The corner points of the feasible region are $(0, 2)$, $(0, 4)$, $(2, 0)$, $(6, 0)$.

2. Values of $Z$: $4$ at $(0, 2)$, $8$ at $(0, 4)$, $2$ at $(2, 0)$, $6$ at $(6, 0)$.

3. The region is bounded, so the minimum is the smallest of these: $2$, at $(2, 0)$.

AnswerA · $(0, 7)$

1. Corners: $(0, 0)$, $(0, 7)$, $(3, 4)$, $(5, 0)$.

2. $Z$ there: $0$, $21$, $18$, $10$; the largest is at $(0, 7)$.