JEE Mathematics Linear Programming MCQs Set 02

Answer: (c) the optimum value of the objective function

Explanation:
1. A linear programming problem seeks the best value of a linear objective function under linear constraints.
2. The best value may be a maximum or a minimum, i.e. the optimum.

Teacher's Note:
1. Optimum covers both maximum and minimum.

Answer: (d) two or more variables

Explanation:
1. Linear constraints may involve two or more variables; the usual graphical case has two.

Teacher's Note:
1. A problem with one variable is not a programming problem.

Answer: (a) the optimal solution of the L.P.P.

Explanation:
1. The optimum of a linear objective over a feasible region always occurs at a corner point (the corner point theorem).

Teacher's Note:
1. Recall the corner point theorem.

Answer: (c) given by the corner points of the feasible region

Explanation:
1. By the corner point theorem the optimum occurs at a vertex of the feasible region, which need not lie on an axis.

Teacher's Note:
1. Optimum at corner points.

Answer: (c) If an L.P.P. admits two optimal solutions, it has an infinite number of optimal solutions

Explanation:
1. A problem may have no optimum (unbounded or infeasible) or many, so (a) and (b) fail.
2. If two points are optimal, every point of the segment joining them (feasible by convexity) gives the same value, so there are infinitely many. The feasible set is convex, so (d) fails.

Teacher's Note:
1. Linear functions are constant-gradient along a segment.

Answer: (b) the half plane containing the point \( (h,k) \) and the points on \( ax+by=4 \)

Explanation:
1. The inequality is not strict, so the boundary line \( ax+by=4 \) is included.
2. The half plane is the one on the same side of the line as \( (h,k) \).

Teacher's Note:
1. A non-strict inequality includes the boundary.

Answer: (c) 35

Explanation:
1. The corner points are \( (0,0) \), \( (7,0) \), \( (4,3) \) and \( (0,5) \).
2. \( z=0,\ 35,\ 26,\ 10 \) respectively, so the maximum is 35 at \( (7,0) \).

Teacher's Note:
1. Evaluate z at every corner point.

Answer: (c) 30

Explanation:
1. The corner points are \( (0,0) \), \( (10,0) \), \( \left(\frac{20}{3},\frac{20}{3}\right) \) and \( (0,10) \).
2. \( P=0,\ 10,\ \frac{80}{3},\ 30 \), so the maximum is 30 at \( (0,10) \).

Teacher's Note:
1. Compute P at each vertex.

Answer: (b) a function to be optimised

Explanation:
1. The objective function is the linear function whose maximum or minimum value is sought.

Teacher's Note:
1. It is the quantity to be optimised.

Answer: (b) \( (0,8) \)

Explanation:
1. \( Z=0,\ 15,\ -2,\ -14,\ -28,\ -32 \) at \( (0,0),(5,0),(6,5),(6,8),(4,10),(0,8) \) respectively.
2. The minimum value \( -32 \) occurs at \( (0,8) \).

Teacher's Note:
1. Evaluate Z at every corner point.

Answer: (a) the first quadrant

Explanation:
1. Both coordinates are non-negative, which describes the first quadrant.

Teacher's Note:
1. Signs of x and y.

Answer: (b) \( (2,5) \)

Explanation:
1. \( Z=24,\ 23,\ 25,\ 36 \) at the four corners.
2. Both coefficients are positive, so Z grows away from the corners in an unbounded region; the minimum is 23 at \( (2,5) \).

Teacher's Note:
1. Positive coefficients keep the minimum at a corner.

Answer: (c) 0

Explanation:
1. The corner points are \( (0,0) \), \( (0,4) \), \( (6,12) \) and \( (6,16) \).
2. \( f=0,\ -16,\ -30,\ -46 \), so the maximum is 0 at \( (0,0) \).

Teacher's Note:
1. List the vertices of the parallelogram.

Answer: (c) 37 at \( (4,5) \)

Explanation:
1. The corner points are \( (0,0) \), \( (7,0) \), \( (6,3) \), \( (4,5) \) and \( (0,6) \).
2. \( Z=0,\ 21,\ 33,\ 37,\ 30 \). The maximum is 37 at \( (4,5) \).

Teacher's Note:
1. Find the intersections of consecutive lines.

Answer: (c) \( \left(2,\dfrac12\right) \)

Explanation:
1. The unbounded feasible region has the corner points \( (0,3) \), \( \left(2,\frac12\right) \) and \( (4,0) \).
2. \( Z=63,\ 22.5,\ 24 \) respectively. The minimum is 22.5 at \( \left(2,\frac12\right) \).

Teacher's Note:
1. Evaluate Z at the corner points.

Answer: (d) 36 at \( (0,6) \)

Explanation:
1. The corner points are \( (4,0) \), \( (0,4) \) and \( (0,6) \).
2. \( Z=16,\ 24,\ 36 \). The maximum is 36 at \( (0,6) \).

Teacher's Note:
1. Draw the strip between the two lines.

Answer: (a) Maximise \( Z=0.08x+0.09y \) subject to \( x+y\le50000 \), \( x\ge15000 \), \( y\ge20000 \)

Explanation:
1. The total invested is at most 50,000, so \( x+y\le50000 \).
2. 'At least' gives \( x\ge15000 \) and \( y\ge20000 \). Income is \( 0.08x+0.09y \).

Teacher's Note:
1. At least translates to a lower bound.

Answer: (b) Maximise \( Z=1000x+1200y \) subject to \( 8x+11y\le450 \), \( 2x+3y\le100 \), \( x,y\ge0 \)

Explanation:
1. Wood used is \( 8x+11y \) and cannot exceed 450. Steel used is \( 2x+3y \) and cannot exceed 100.
2. So both constraints are '\( \le \)', giving option (b).

Teacher's Note:
1. Resource usage cannot exceed availability.

Answer: (d) any point on the segment joining \( (0,2) \) and \( (3,0) \)

Explanation:
1. \( F(0,2)=12 \) and \( F(3,0)=12 \), and the other corners give 24, 84 and 30.
2. The minimum 12 is attained at two corners, so it is attained at every point of the segment joining them.

Teacher's Note:
1. Equal values at two corners mean a whole edge of optimal points.

Answer: (c) \( (2,0) \)

Explanation:
1. The corners are \( (0,0) \), \( (2,0) \) and \( (0,2) \), with values 0, 6 and 4.
2. The maximum is at \( (2,0) \).

Teacher's Note:
1. Evaluate at each vertex.

Answer: (d) none of these

Explanation:
1. The lines meet at \( (3,0) \); the region is the triangle with vertices \( (0,0) \), \( (3,0) \) and \( (0,2) \).
2. All three listed points are vertices, so the answer is 'none of these'.

Teacher's Note:
1. Find the intersection of the two lines first.

Answer: (b) 120

Explanation:
1. The corner points are \( (0,0) \), \( (15,0) \), \( (12,6) \) and \( (0,12) \).
2. \( z=0,\ 90,\ 120,\ 96 \). The maximum is 120.

Teacher's Note:
1. The vertex \( (12,6) \) is the intersection of the two lines.