मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Minimize Z = x + 4y subject to constraints x + 3y ≥ 3, 2x + y ≥ 2, x ≥ 0, y ≥ 0 - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Minimize Z = x + 4y subject to constraints

x + 3y ≥ 3, 2x + y ≥ 2, x ≥ 0, y ≥ 0

तक्ता
आकृती
Advertisements

उत्तर

To draw the feasible region, construct table as follows:

Inequality x + 3y ≥ 3 2x + y ≥ 2
Corresponding equation (of line) x + 3y = 3 2x + y = 2
Intersection of line with X-axis (3, 0) (1, 0)
Intersection of line with Y-axis (0, 1) (0, 2)
Region Non-origin side Non-origin side

Shaded portion XABCY is the feasible region, whose vertices are A(3, 0), B and C(0, 2).

B is the point of intersection of the lines 2x + y = 2 and x + 3y = 3.

∴ B ≡ `(3/5, 4/5)`


Here, the objective function is

Z = x + 4y

∴ Z at A(3, 0) = 3 + 4(0)

= 3

Z at B`(3/5, 4/5) = 3/5 + 4(4/5)`

= `19/5`

= 3.8

Z at C(0, 2) = 0 + 4(2)

= 8

∴ Z has minimum value 3 at x = 3 and y = 0.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2.6: Linear Programming - Q.4 (D)

संबंधित प्रश्‍न

A firm manufactures 3 products AB 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.


Amit's mathematics teacher has given him three very long lists of problems with the instruction to submit not more than 100 of them (correctly solved) for credit. The problem in the first set are worth 5 points each, those in the second set are worth 4 points each, and those in the third set are worth 6 points each. Amit knows from experience that he requires on the average 3 minutes to solve a 5 point problem, 2 minutes to solve a 4 point problem, and 4 minutes to solve a 6 point problem. Because he has other subjects to worry about, he can not afford to devote more than

\[3\frac{1}{2}\] hours altogether to his mathematics assignment. Moreover, the first two sets of problems involve numerical calculations and he knows that he cannot stand more than 
\[2\frac{1}{2}\]  hours work on this type of problem. Under these circumstances, how many problems in each of these categories shall he do in order to get maximum possible credit for his efforts? Formulate this as a LPP.

 


A firm manufactures two products, each of which must be processed through two departments, 1 and 2. The hourly requirements per unit for each product in each department, the weekly capacities in each department, selling price per unit, labour cost per unit, and raw material cost per unit are summarized as follows:
 

  Product A Product B Weekly capacity
Department 1 3 2 130
Department 2 4 6 260
Selling price per unit ₹ 25 ₹ 30  
Labour cost per unit ₹ 16 ₹ 20  
Raw material cost per unit ₹ 4 ₹ 4  


The problem is to determine the number of units to produce each product so as to maximize total contribution to profit. Formulate this as a LPP.


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.


Choose the correct alternative :

The maximum value of z = 5x + 3y. subject to the constraints


Choose the correct alternative :

The point at which the maximum value of z = x + y subject to the constraints x + 2y ≤ 70, 2x + y ≤ 95, x ≥ 0, y ≥ 0 is


Fill in the blank :

The region represented by the in equations x ≤ 0, y ≤ 0 lines in _______ quadrants.


The constraint that a factory has to employ more women (y) than men (x) is given by _______


Solve the following problem :

Maximize Z = 5x1 + 6x2 Subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x ≥ 0, x2 ≥ 0


Solve the following problem:

Maximize Z = 4x1 + 3x2 Subject to 3x1 + x2 ≤ 15, 3x1 + 4x2 ≤ 24, x1 ≥ 0, x2 ≥ 0


Maximize Z = 60x + 50y Subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0


Solve the following problem :

A factory produced two types of chemicals A and B The following table gives the units of ingredients P & Q (per kg) of Chemicals A and B as well as minimum requirements of P and Q and also cost per kg. of chemicals A and B.

Ingredients per kg. /Chemical Units A
(x)
B
(y)
Minimum requirements in
P 1 2 80
Q 3 1 75
Cost (in ₹) 4 6  

Find the number of units of chemicals A and B should be produced so as to minimize the cost.


Choose the correct alternative:

The maximum value of Z = 3x + 5y subjected to the constraints x + y ≤ 2, 4x + 3y ≤ 12, x ≥ 0, y ≥ 0 is


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:

Of all the points of feasible region, the optimal value is obtained at the boundary of the feasible region


A dealer deals in two products X and Y. He has ₹ 1,00,000/- to invest and space to store 80 pieces. Product X costs ₹ 2500/- and product Y costs ₹ 1000/- per unit. He can sell the items X and Y at respective profits of ₹ 300 and ₹ 90. Construct the LPP and find the number of units of each product to be purchased to maximize its profit


A chemist has a compound to be made using 3 basic elements X, Y, Z so that it has at least 10 litres of X, 12 litres of Y and 20 litres of Z. He makes this compound by mixing two compounds (I) and (II). Each unit compound (I) had 4 litres of X, 3 litres of Y. Each unit compound (II) had 1 litre of X, 2 litres of Y and 4 litres of Z. The unit costs of compounds (I) and (II) are ₹ 400 and ₹ 600 respectively. Find the number of units of each compound to be produced so as to minimize the cost


Maximize Z = 5x + 10y subject to constraints

x + 2y ≤ 10, 3x + y ≤ 12, x ≥ 0, y ≥ 0


Maximize Z = 400x + 500y subject to constraints

x + 2y ≤ 80, 2x + y ≤ 90, x ≥ 0, y ≥ 0


Amartya wants to invest ₹ 45,000 in Indira Vikas Patra (IVP) and in Public Provident fund (PPF). He wants to invest at least ₹ 10,000 in PPF and at least ₹ 5000 in IVP. If the rate of interest on PPF is 8% per annum and that on IVP is 7% per annum. Formulate the above problem as LPP to determine maximum yearly income.

Solution: Let x be the amount (in ₹) invested in IVP and y be the amount (in ₹) invested in PPF.

x ≥ 0, y ≥ 0

As per the given condition, x + y ______ 45000

He wants to invest at least ₹ 10,000 in PPF.

∴ y ______ 10000

Amartya wants to invest at least ₹ 5000 in IVP.

∴ x ______ 5000

Total interest (Z) = ______

The formulated LPP is

Maximize Z = ______ subject to 

______


Solve the following LPP graphically:

Maximize Z = 9x + 13y subject to constraints

2x + 3y ≤ 18, 2x + y ≤ 10, x ≥ 0, y ≥ 0

Solution: Convert the constraints into equations and find the intercept made by each one of it.

Inequation Equation X intercept Y intercept Region
2x + 3y ≤ 18 2x + 3y = 18 (9, 0) (0, ___) Towards origin
2x + y ≤ 10 2x + y = 10 ( ___, 0) (0, 10) Towards origin
x ≥ 0, y ≥ 0 x = 0, y = 0 X axis Y axis ______

The feasible region is OAPC, where O(0, 0), A(0, 6),

P( ___, ___ ), C(5, 0)

The optimal solution is in the following table:

Point Coordinates Z = 9x + 13y Values Remark
O (0, 0) 9(0) + 13(0) 0  
A (0, 6) 9(0) + 13(6) ______  
P ( ___,___ ) 9( ___ ) + 13( ___ ) ______ ______
C (5, 0) 9(5) + 13(0) ______  

∴ Z is maximum at __( ___, ___ ) with the value ___.


Solve the LPP graphically:
Minimize Z = 4x + 5y
Subject to the constraints 5x + y ≥ 10, x + y ≥ 6, x + 4y ≥ 12, x, y ≥ 0

Solution: Convert the constraints into equations and find the intercept made by each one of it.

Inequations Equations X intercept Y intercept Region
5x + y ≥ 10 5x + y = 10 ( ___, 0) (0, 10) Away from origin
x + y ≥ 6 x + y = 6 (6, 0) (0, ___ ) Away from origin
x + 4y ≥ 12 x + 4y = 12 (12, 0) (0, 3) Away from origin
x, y ≥ 0 x = 0, y = 0 x = 0 y = 0 1st quadrant

∵ Origin has not satisfied the inequations.

∴ Solution of the inequations is away from origin.

The feasible region is unbounded area which is satisfied by all constraints.

In the figure, ABCD represents

The set of the feasible solution where

A(12, 0), B( ___, ___ ), C ( ___, ___ ) and D(0, 10).

The coordinates of B are obtained by solving equations

x + 4y = 12 and x + y = 6

The coordinates of C are obtained by solving equations

5x + y = 10 and x + y = 6

Hence the optimum solution lies at the extreme points.

The optimal solution is in the following table:

Point Coordinates Z = 4x + 5y Values Remark
A (12, 0) 4(12) + 5(0) 48  
B ( ___, ___ ) 4( ___) + 5(___ ) ______ ______
C ( ___, ___ ) 4( ___) + 5(___ ) ______  
D (0, 10) 4(0) + 5(10) 50  

∴ Z is minimum at ___ ( ___, ___ ) with the value ___


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:-


Maximised value of z in z = 3x + 4y, subject to constraints : x + y ≤ 4, x ≥ 0. y ≥ 0


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×