OP Malhotra Class 12 Maths Solutions Chapter 28 Linear Programming Exercise 28 (B)

S Chand Class 12 ICSE Maths Solutions Chapter 28 Linear Programming Ex 28(b)

 

Question 1. Find the maximum value of each expression on region R
(i) \( 3x + 4y \)
(ii) \( 2y - x \)

Answer:

The feasible region R is a polygon with corner points: O(0,0), A(0,4), B(3,4), C(5,2), and D(6,0). This region is typically found by graphing a set of inequalities.

(i) To find the maximum value of \( Z = 3x + 4y \):
From the table, the maximum value of \( Z \) is 25, which occurs at point B (3, 4). This means when \( x=3 \) and \( y=4 \), the expression reaches its highest value within the allowed region.
(ii) To find the maximum value of \( Z = 2y - x \):
From this evaluation, the maximum value of \( Z \) is 8, occurring at point A (0, 4). Linear programming problems always find optimal values at the vertices of the feasible region.In simple words: We looked at the special points (corners) of the shaded area. For each expression, we put the x and y values from these corners into the formula to see which one gave the biggest answer. The highest number we got for the first expression was 25, and for the second expression, it was 8.

🎯 Exam Tip: Always evaluate the objective function at all corner points of the feasible region. Clearly mark the maximum or minimum value in your table for easy identification by the examiner.

 

Question 2. Find the minimum value of each expression on region S shown in
(i) \( x + y \)
(ii) \( 5x + 4y \)

Answer:

The feasible region S is a polygon defined by the corner points O(0,0), E(0,10), F(5,2), and G(7,0). This region represents all possible solutions that satisfy the given constraints.

(i) To find the minimum value of \( Z = x + y \):
The minimum value of \( Z \) is 7, which occurs along the line segment joining points F (5, 2) and G (7, 0). This happens when multiple points give the same optimal value.
(ii) To find the minimum value of \( Z = 5x + 4y \):
Here, the minimum value of \( Z \) is 33, occurring at point F (5, 2). The objective function changes its optimal point depending on the coefficients.In simple words: We checked the same corner points again, but this time for a different expression. For the first one (\( x+y \)), the smallest value was 7, found at two points on a line. For the second one (\( 5x+4y \)), the smallest value was 33, found at just one corner point.

🎯 Exam Tip: If the minimum or maximum value occurs at two corner points, it means all points on the line segment connecting those two points also yield the same optimal value.

 

Question 3.
(i) Evaluate the linear expresion \( 3x + 2y \) at each corner point of the convex polygonal region shown in the graph below :
(ii) What is the minimum value of \( 3x + 2y \) over the region shown in the graph above ?

Answer:

The feasible region for this problem is an unbounded region in the first quadrant, defined by the corner points (0,7), (2,3), (4,1), and (8,0). These points form the boundary of the region that extends infinitely upwards and to the right.

(i) Evaluating \( Z = 3x + 2y \) at each corner point:
(ii) The minimum value of \( Z = 3x + 2y \) is 12, which occurs at the corner point (2, 3). For unbounded feasible regions, the minimum or maximum value may or may not exist, but if it does, it will be at a corner point.In simple words: We looked at the corner points of the graph and put their \( x \) and \( y \) numbers into the formula \( 3x + 2y \). The smallest answer we got was 12, which happened at the point (2,3). This region goes on forever in some directions, but the lowest point for our expression was still at a corner.

🎯 Exam Tip: For unbounded regions, after finding the minimum or maximum value at a corner point, it's good practice to check if the region defined by \( Z < Z_{min} \) (for minimum) or \( Z > Z_{max} \) (for maximum) has any common points with the feasible region. If not, the value found is indeed the optimum.

 

Question 4. Refer to the following system of inequalities \( x \ge 1 \), \( x \ge 2 \), \( 3y + x \ge 9 \), \( y + 2x \ge 8 \)
(i) Graph the solution set of the system.
(ii) Determine the corner points of the region that represents the solution set.
(iii) Find the minimum value of each of the following linear expresions over the region graphed in (i) by evaluating the expression at each of the corner points.
(a) \( x + 5y \)
(b) \( 5x + y \)
(c) \( x + y \)

Answer:

First, we convert the given inequalities into equations to find the boundary lines:
\( x = 2 \) (combining \( x \ge 1 \) and \( x \ge 2 \))
\( 3y + x = 9 \)
\( y + 2x = 8 \)
The solution also implies an additional constraint \( y \ge 1 \) which, along with these, defines the feasible region. We also have \( x \ge 0 \) and \( y \ge 0 \) for the first quadrant.

The corner points of the unbounded feasible region, found by intersecting these lines, are:
P (6,1) (intersection of \( y=1 \) and \( 3y+x=9 \))
Q (3,2) (intersection of \( 3y+x=9 \) and \( y+2x=8 \))
R (2,4) (intersection of \( x=2 \) and \( y+2x=8 \))

(iii) Evaluating the expressions at the corner points:
(a) For \( Z = x + 5y \):
The minimum value of \( Z = x + 5y \) is 11.
(b) For \( Z = 5x + y \):
The minimum value of \( Z = 5x + y \) is 14.
(c) For \( Z = x + y \):
The minimum value of \( Z = x + y \) is 5. It is important to remember that optimal solutions for these problems are found at the 'corners' or vertices of the feasible region.In simple words: We first figured out the boundaries of the allowed area by drawing lines for all the rules. This area was unbounded, meaning it stretched out infinitely in some directions, but had specific corner points: P(6,1), Q(3,2), and R(2,4). Then, we tested each of the three expressions at these corner points to find the smallest value for each. For \( x+5y \), the minimum was 11; for \( 5x+y \), it was 14; and for \( x+y \), it was 5.

🎯 Exam Tip: When dealing with multiple inequalities, carefully identify all intersection points that form the vertices of the feasible region. For unbounded regions, verify the optimality of the corner point by checking if the objective function can be further improved in the unbounded direction.

 

Question 5. Refer to the following system of inequalities. \( x \ge 1 \), \( y \le 6 \), \( y \ge -x + 4 \), \( 2y \ge x - 1 \), \( 3y \le -x + 21 \)
(i) Graph the solution set of the system and label the corner points with their coordinates,
(ii) Find the minimum value of the expression \( x + 5y \) over the region graphed in (i).
(iii) Find the maximum value of the expression \( 4x - 3y \) over the region graphed in (i).

Answer:

First, we convert the inequalities into equations to find the boundary lines:
\( x = 1 \)
\( y = 6 \)
\( y = -x + 4 \) (or \( x + y = 4 \))
\( 2y = x - 1 \) (or \( x - 2y = 1 \))
\( 3y = -x + 21 \) (or \( x + 3y = 21 \))
We also consider \( x \ge 0 \) and \( y \ge 0 \) for the first quadrant.

The corner points of the bounded feasible region are found by intersecting these lines:
P (3,1) (intersection of \( x+y=4 \) and \( x-2y=1 \))
Q (9,4) (intersection of \( x-2y=1 \) and \( x+3y=21 \))
T (3,6) (intersection of \( y=6 \) and \( x+3y=21 \))
S (1,6) (intersection of \( x=1 \) and \( y=6 \))
R (1,3) (intersection of \( x=1 \) and \( x+y=4 \))

(ii) To find the minimum value of \( Z = x + 5y \):
The minimum value of \( Z = x + 5y \) is 8.
(iii) To find the maximum value of \( Z = 4x - 3y \):
The maximum value of \( Z = 4x - 3y \) is 24.In simple words: First, we drew the lines for all the rules to find the allowed area, which was a specific shape with five corners. Then, we calculated the value of two different expressions at each of these five corners. For the first expression, the smallest value was 8. For the second expression, the largest value was 24. These values are the best possible results within the allowed region.

🎯 Exam Tip: Always list all inequalities, derive the equations, find all intersection points, and clearly identify which ones form the vertices of the feasible region. A clear graph helps in identifying the correct corner points, especially for complex regions.

 

Question 6. Minimize \( Z = 20x + 10y \) subject to the constraints \( x + 2y \le 40 \), \( 3x + y \ge 30 \), \( 4x + 3y \ge 60 \), \( x \ge 0 \); \( y \ge 0 \)

Answer:

First, we convert the inequalities into equations to define the boundary lines:
\( x + 2y = 40 \)
\( 3x + y = 30 \)
\( 4x + 3y = 60 \)
And the non-negativity constraints are \( x = 0 \) and \( y = 0 \).

The feasible region is a bounded polygon in the first quadrant, defined by the following corner points:
E (15,0) (intersection of \( 4x+3y=60 \) and \( y=0 \))
A (40,0) (intersection of \( x+2y=40 \) and \( y=0 \))
P (6,12) (intersection of \( 3x+y=30 \) and \( 4x+3y=60 \))
Q (4,18) (intersection of \( x+2y=40 \) and \( 3x+y=30 \))

Now, we evaluate \( Z = 20x + 10y \) at each of these corner points:
The minimum value of \( Z \) is 240, which occurs at point P (6, 12). Finding the feasible region correctly is the most important step in these problems.In simple words: We first drew the graph for all the rules to find the allowed area, which was a four-sided shape with corners E(15,0), A(40,0), P(6,12), and Q(4,18). Then, we put the x and y values from each corner into the expression \( 20x + 10y \). The smallest answer we got was 240, which happened at point P(6,12).

🎯 Exam Tip: Always solve the systems of equations carefully to find the exact coordinates of the corner points. Even a small error in coordinates can lead to an incorrect optimal value.

 

Question 7. Minimize \( Z = 3x + 4y \) subject to constraints \( x + y \ge 3 \), \( 2x + y \ge 4 \), \( x, y \ge 0 \)

Answer:

First, we convert the inequalities into equations for the boundary lines:
\( x + y = 3 \)
\( 2x + y = 4 \)
And the non-negativity constraints are \( x = 0 \) and \( y = 0 \).

The corner points of the feasible region, which is unbounded, are:
A (3,0) (intersection of \( x+y=3 \) and \( y=0 \))
P (1,2) (intersection of \( x+y=3 \) and \( 2x+y=4 \))
D (0,4) (intersection of \( 2x+y=4 \) and \( x=0 \))

Now, we evaluate \( Z = 3x + 4y \) at each of these corner points:
The smallest value of \( Z \) found at a corner point is 9, at A (3, 0). Since the feasible region is unbounded, we must check if this minimum is actual. We consider the open half-plane \( 3x + 4y < 9 \). If this half-plane has no points in common with the feasible region, then 9 is indeed the minimum value. In this case, there are no common points, so \( Z_{min} = 9 \) at \( x = 3 \) and \( y = 0 \).In simple words: We drew the lines for the rules to find the allowed area, which was an unbounded region with three corners: A(3,0), P(1,2), and D(0,4). We then put the \( x \) and \( y \) values from these corners into the expression \( 3x + 4y \). The smallest value we found was 9 at point A(3,0). Because the region is unbounded, we had to do an extra check to make sure this really was the smallest possible value, and it was.

🎯 Exam Tip: For unbounded feasible regions, after calculating the objective function at corner points, always perform the additional test: draw the half-plane corresponding to \( Z < Z_{min} \) (for minimization) or \( Z > Z_{max} \) (for maximization). If this half-plane does not overlap with the feasible region, the calculated optimal value is correct.

 

Question 8. Maximize \( Z = 6x_1 + 11x_2 \) subject to the constraints \( 2x_1 + x_2 \le 104 \), \( x_1 + 2x_2 \le 76 \), \( x_1, x_2 \ge 0 \)

Answer:

First, we convert the inequalities into equations to find the boundary lines:
\( 2x_1 + x_2 = 104 \)
\( x_1 + 2x_2 = 76 \)
And the non-negativity constraints are \( x_1 = 0 \) and \( x_2 = 0 \).

The corner points of the bounded feasible region are:
O (0,0)
A (52,0) (intersection of \( 2x_1+x_2=104 \) and \( x_2=0 \))
P (44,16) (intersection of \( 2x_1+x_2=104 \) and \( x_1+2x_2=76 \))
D (0,38) (intersection of \( x_1+2x_2=76 \) and \( x_1=0 \))

Now, we evaluate \( Z = 6x_1 + 11x_2 \) at each of these corner points:
The maximum value of \( Z \) is 440, which occurs at point P (44, 16). The optimal solution always lies at a vertex of the feasible region for bounded problems.In simple words: We first set up the boundaries using the given rules and found that the allowed area was a four-sided shape with corners at O(0,0), A(52,0), P(44,16), and D(0,38). We then put the \( x_1 \) and \( x_2 \) values from each corner into the expression \( 6x_1 + 11x_2 \). The biggest answer we got was 440, which was at point P(44,16).

🎯 Exam Tip: When dealing with multiple variables like \( x_1 \) and \( x_2 \), ensure consistency in notation throughout your solution. Clearly labelling the axes and corner points on your graph can help prevent errors.