Advertisements
Advertisements
प्रश्न
Smita is a diet conscious house wife, wishes to ensure certain minimum intake of vitamins A, B and C for the family. The minimum daily needs of vitamins A, B, and C for the family are 30, 20, and 16 units respectively. For the supply of the minimum vitamin requirements Smita relies on 2 types of foods F1 and F2. F1 provides 7, 5 and 2 units of A, B, C vitamins per 10 grams and F2 provides 2, 4 and 8 units of A, B and C vitamins per 10 grams. F1 costs ₹ 3 and F2 costs ₹ 2 per 10 grams. How many grams of each F1 and F2 should buy every day to keep her food bill minimum
उत्तर
Let food F1 be x grams and food F2 be y grams.
Since x and y cannot be negative, x ≥ 0, y ≥ 0.
F1 costs ₹ 3 and F2 costs ₹ 2 per 10 grams.
∴ Total cost = Z = 3x + 2y
We construct a table with constraints of vitamins A, B and C as follows:
Vitamins/Food | F1 | F2 | Minimum requirement |
A | 7 | 2 | 30 |
B | 5 | 4 | 20 |
C | 2 | 8 | 16 |
From the table, the constraints are
7x + 2y ≥ 30
5x + 4y ≥ 20
2x + 8y ≥ 16
∴ Given problem can be formulated as follows:
Minimize Z = 3x + 2y
Subject to 7x + 2y ≥ 30
5x + 4y ≥ 20,
2x + 8y ≥ 16, x ≥ 0, y ≥ 0
To draw the feasible region, construct table as follows:
Inequality | 7x + 2y ≥ 30 | 5x + 4y ≥ 20 | 2x + 8y ≥ 16 |
Corresponding equation (of line) | 7x + 2y = 30 | 5x + 4y = 20 | 2x + 8y = 16 |
Intersection of line with X-axis | `(30/7, 0)` | (4, 0) | (8, 0) |
Intersection of line with Y-axis | (0, 15) | (0, 5) | (0, 2) |
Region | Non-Origin side | Non-Origin side | Non-Origin side |
Shaded portion XABCY is the feasible region, whose vertices are A(8, 0), B and C(0, 15),
B is the point of intersection of the lines 7x + 2y = 30 and 2x + 8y = 16
Solving the above equations, we get
x = 4, y = 1
∴ B ≡ (4, 1)
Here, the objective function is
Z = 3x + 2y
Z at A (8, 0) = 3(8) + 2(0) = 24
Z at B (4, 1) = 3(4) + 2(1)
= 12 + 2
= 14
Z at C (0, 15) = 3(0) + 2(15)
= 30
∴ Z has minimum value 14 at x = 4 and y = 1.
∴ Smita should buy 4 grams of food F1 and 1 gram of food F2 every day to keep her food bill minimum.
संबंधित प्रश्न
A company produces two types of goods A and B, that require gold and silver. Each unit of type A requires 3 g of silver and 1 g of golds while that of type B requires 1 g of silver and 2 g of gold. The company can procure a maximum of 9 g of silver and 8 g of gold. If each unit of type A brings a profit of Rs 40 and that of type B Rs 50, formulate LPP to maximize profit.
A firm manufactures 3 products A, B and C. The profits are Rs 3, Rs 2 and Rs 4 respectively. The firm has 2 machines and below is the required processing time in minutes for each machine on each product :
Machine | Products | ||
A | B | C | |
M1 M2 |
4 | 3 | 5 |
2 | 2 | 4 |
Machines M1 and M2 have 2000 and 2500 machine minutes respectively. The firm must manufacture 100 A's, 200 B's and 50 C's but not more than 150 A's. Set up a LPP to maximize the profit.
Solve the following LPP by graphical method:
Maximize z = 11x + 8y, subject to x ≤ 4, y ≤ 6, x + y ≤ 6, x ≥ 0, y ≥ 0
Solve the following L.P.P. by graphical method :
Maximize : Z = 7x + 11y subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Solve the following L.P.P. by graphical method :
Maximize: Z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0 also find maximum value of Z.
Fill in the blank :
Graphical solution set of the in equations x ≥ 0, y ≥ 0 is in _______ quadrant
Fill in the blank :
The region represented by the in equations x ≤ 0, y ≤ 0 lines in _______ quadrants.
The region represented by the inequality y ≤ 0 lies in _______ quadrants.
The constraint that a factory has to employ more women (y) than men (x) is given by _______
The region represented by the inequalities x ≥ 0, y ≥ 0 lies in first quadrant.
Solve the following problem :
Maximize Z = 5x1 + 6x2 Subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x ≥ 0, x2 ≥ 0
Solve the following problem :
Minimize Z = 2x + 3y Subject to x – y ≤ 1, x + y ≥ 3, x ≥ 0, y ≥ 0
Solve the following problem :
Maximize Z = 4x1 + 3x2 Subject to 3x1 + x2 ≤ 15, 3x1 + 4x2 ≤ 24, x1 ≥ 0, x2 ≥ 0
Solve the following problem :
Maximize Z = 60x + 50y Subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0
Choose the correct alternative:
The point at which the maximum value of Z = 4x + 6y subject to the constraints 3x + 2y ≤ 12, x + y ≥ 4, x ≥ 0, y ≥ 0 is obtained at the point
Choose the correct alternative:
The corner points of feasible region for the inequations, x + y ≤ 5, x + 2y ≤ 6, x ≥ 0, y ≥ 0 are
Choose the correct alternative:
The corner points of the feasible region are (4, 2), (5, 0), (4, 1) and (6, 0), then the point of minimum Z = 3.5x + 2y = 16 is at
State whether the following statement is True or False:
The maximum value of Z = 5x + 3y subjected to constraints 3x + y ≤ 12, 2x + 3y ≤ 18, 0 ≤ x, y is 20
State whether the following statement is True or False:
A convex set includes the points but not the segment joining the points
State whether the following statement is True or False:
If the corner points of the feasible region are (0, 10), (2, 2) and (4, 0), then the minimum value of Z = 3x + 2y is at (4, 0)
State whether the following statement is True or False:
Corner point method is most suitable method for solving the LPP graphically
State whether the following statement is True or False:
Of all the points of feasible region, the optimal value is obtained at the boundary of the feasible region
The feasible region represented by the inequations x ≥ 0, y ≤ 0 lies in ______ quadrant.
A company manufactures two types of ladies dresses C and D. The raw material and labour available per day is given in the table.
Resources | Dress C(x) | Dress D(y) | Max. availability |
Raw material | 5 | 4 | 60 |
Labour | 5 | 3 | 50 |
P is the profit, if P = 50x + 100y, solve this LPP to find x and y to get the maximum profit
Maximize Z = 2x + 3y subject to constraints
x + 4y ≤ 8, 3x + 2y ≤ 14, x ≥ 0, y ≥ 0.
A linear function z = ax + by, where a and b are constants, which has to be maximised or minimised according to a set of given condition is called a:-