Advertisements
Advertisements
Question
Maximize Z = 7x + 10y
Subject to
\[x + y \leq 30000\]
\[ y \leq 12000\]
\[ x \geq 6000\]
\[ x \geq y\]
\[ x, y \geq 0\]
Advertisements
Solution
We have to maximize Z = 7x + 10y
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 30000,y = 12000, x = 6000, x = y, x = 0 and y = 0.
Region represented by x + y ≤ 30000:
The line x + y = 30000 meets the coordinate axes at \[A\left( 30000, 0 \right)\] and \[B\left( 0, 30000 \right)\] respectively. By joining these points we obtain the line x + y = 30000.
Clearly (0,0) satisfies the inequation x + y ≤ 30000. So,the region containing the origin represents the solution set of the inequation x + y ≤ 30000.
The line y = 12000 is the line that passes through C(0,12000) and parallel to x axis.
The line x = 6000 is the line that passes through (6000, 0) and parallel to y axis.
Region represented by x ≥ y
The line x = y is the line that passes through origin.The points to the right of the line x= y satisfy the inequation x ≥ y.
Like by taking the point (−12000, 6000).Here, 6000 > −12000 which implies y > x. Hence, the points to the left of the line x = y will not satisfy the given inequation x ≥ y.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
The feasible region determined by the system of constraints, x + y ≤ 30000, y ≤ 12000, x ≥ 6000, x ≥ y , x ≥ 0 and y ≥ 0 are as follows:
The corner points of the feasible region are D(6000, 0), \[A\left( 3000, 0 \right)\] , \[F\left( 18000, 12000 \right)\] and \[E\left( 12000, 12000 \right)\] .
The values of Z at these corner points are as follows:
| Corner point | Z = 7x + 10y |
| D(6000, 0) | 7 × 6000 + 10 × 0 = 42000 |
|
\[A\left( 3000, 0 \right)\]
|
7× 3000 + 10 × 0 = 21000 |
|
\[F\left( 18000, 12000 \right)\]
|
7 × 18000 + 10 × 12000 = 246000 |
|
\[E\left( 12000, 12000 \right)\]
|
7 × 12000 + 10 ×12000 = 204000 |
We see that the maximum value of the objective function Z is 246000 which is at \[F\left( 18000, 12000 \right)\] that means at x = 18000 and y = 12000.
Thus, the optimal value of Z is 246000.
APPEARS IN
RELATED QUESTIONS
Minimize: Z = 6x + 4y
Subject to the conditions:
3x + 2y ≥ 12,
x + y ≥ 5,
0 ≤ x ≤ 4,
0 ≤ y ≤ 4
Minimize :Z=6x+4y
Subject to : 3x+2y ≥12
x+y ≥5
0 ≤x ≤4
0 ≤ y ≤ 4
There are two types of fertilisers 'A' and 'B'. 'A' consists of 12% nitrogen and 5% phosphoric acid whereas 'B' consists of 4% nitrogen and 5% phosphoric acid. After testing the soil conditions, farmer finds that he needs at least 12 kg of nitrogen and 12 kg of phosphoric acid for his crops. If 'A' costs Rs 10 per kg and 'B' cost Rs 8 per kg, then graphically determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost
A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30, respectively. The company makes a profit of Rs 80 on each piece of type A and Rs 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?
Solve the following L.P.P. graphically Maximise Z = 4x + y
Subject to following constraints x + y ≤ 50
3x + y ≤ 90,
x ≥ 10
x, y ≥ 0
Solve the following L.P.P graphically: Maximise Z = 20x + 10y
Subject to the following constraints x + 2y ≤ 28,
3x + y ≤ 24,
x ≥ 2,
x, y ≥ 0
A dietician wishes to mix two kinds ·of food X· and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B arid 8 units of vitamin C. The vitamin contents of one kg food is given below:
| Food | Vitamin A | Vitamin.B | Vitamin C |
| X | 1 unit | 2 unit | 3 unit |
| Y | 2 unit | 2 unit | 1 unit |
Orie kg of food X costs Rs 24 and one kg of food Y costs Rs 36. Using Linear Programming, find the least cost of the total mixture. which will contain the required vitamins.
Maximize Z = 15x + 10y
Subject to
\[3x + 2y \leq 80\]
\[2x + 3y \leq 70\]
\[ x, y \geq 0\]
Maximize Z = 10x + 6y
Subject to
\[3x + y \leq 12\]
\[2x + 5y \leq 34\]
\[ x, y \geq 0\]
Maximize Z = −x1 + 2x2
Subject to
\[- x_1 + 3 x_2 \leq 10\]
\[ x_1 + x_2 \leq 6\]
\[ x_1 - x_2 \leq 2\]
\[ x_1 , x_2 \geq 0\]
One kind of cake requires 300 gm of flour and 15 gm of fat, another kind of cake requires 150 gm of flour and 30 gm of fat. Find the maximum number of cakes which can be made from 7.5 kg of flour and 600 gm of fat, assuming that there is no shortage of the other ingradients used in making the cake. Make it as an LPP and solve it graphically.
One kind of cake requires 200 g of flour and 25 g of fat, and another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no storage of the other ingredients used in making the cakes.
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:
| Food | Vitamin A | Vitamin B | Vitamin C |
| X | 1 | 2 | 3 |
| Y | 2 | 2 | 1 |
One kg of food X costs ₹16 and one kg of food Y costs ₹20. Find the least cost of the mixture which will produce the required diet?
A company produces two types of leather belts, say type A and B. Belt A is a superior quality and belt B is of a lower quality. Profits on each type of belt are Rs 2 and Rs 1.50 per belt, respectively. Each belt of type A requires twice as much time as required by a belt of type B. If all belts were of type B, the company could produce 1000 belts per day. But the supply of leather is sufficient only for 800 belts per day (both A and B combined). Belt A requires a fancy buckle and only 400 fancy buckles are available for this per day. For belt of type B, only 700 buckles are available per day.
How should the company manufacture the two types of belts in order to have a maximum overall profit?
A firm manufactures two products A and B. Each product is processed on two machines M1 and M2. Product A requires 4 minutes of processing time on M1 and 8 min. on M2 ; product B requires 4 minutes on M1 and 4 min. on M2. The machine M1 is available for not more than 8 hrs 20 min. while machine M2 is available for 10 hrs. during any working day. The products A and B are sold at a profit of Rs 3 and Rs 4 respectively.
Formulate the problem as a linear programming problem and find how many products of each type should be produced by the firm each day in order to get maximum profit.
A manufacturer produces two types of steel trunks. He has two machines A and B. For completing, the first types of the trunk requires 3 hours on machine A and 3 hours on machine B, whereas the second type of the trunk requires 3 hours on machine A and 2 hours on machine B. Machines A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of Rs 30 and Rs 25 per trunk of the first type and the second type respectively. How many trunks of each type must he make each day to make maximum profit?
A gardener has supply of fertilizer of type I which consists of 10% nitrogen and 6% phosphoric acid and type II fertilizer which consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, he finds that he needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for his crop. If the type I fertilizer costs 60 paise per kg and type II fertilizer costs 40 paise per kg, determine how many kilograms of each fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Anil wants to invest at most Rs 12000 in Saving Certificates and National Saving Bonds. According to rules, he has to invest at least Rs 2000 in Saving Certificates and at least Rs 4000 in National Saving Bonds. If the rate of interest on saving certificate is 8% per annum and the rate of interest on National Saving Bond is 10% per annum, how much money should he invest to earn maximum yearly income? Find also his maximum yearly income.
A box manufacturer makes large and small boxes from a large piece of cardboard. The large boxes require 4 sq. metre per box while the small boxes require 3 sq. metre per box. The manufacturer is required to make at least three large boxes and at least twice as many small boxes as large boxes. If 60 sq. metre of cardboard is in stock, and if the profits on the large and small boxes are Rs 3 and Rs 2 per box, how many of each should be made in order to maximize the total profit?
If a young man drives his vehicle at 25 km/hr, he has to spend ₹2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to ₹5 per km. He has ₹100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same.
A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs 25,000 and Rs 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and his profit on the desktop model is Rs 4500 and on the portable model is Rs 5000. Make an LPP and solve it graphically.
There are two factories located one at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are respectively 8 and 6 units. The cost of transportation per unit is given below:
| From \ To | Cost (in ₹) | ||
| A | B | C | |
| P | 160 | 100 | 150 |
| Q | 100 | 120 | 100 |
How many units should be transported from each factory to each depot in order that the transportation cost is minimum. What will be the minimum transportation cost?
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:
| Types of Toys | Machines | ||
| I | II | III | |
| A | 12 | 18 | 6 |
| B | 6 | 0 | 9 |
A carpenter has 90, 80 and 50 running feet respectively of teak wood, plywood and rosewood which is used to product A and product B. Each unit of product A requires 2, 1 and 1 running feet and each unit of product B requires 1, 2 and 1 running feet of teak wood, plywood and rosewood respectively. If product A is sold for Rs. 48 per unit and product B is sold for Rs. 40 per unit, how many units of product A and product B should be produced and sold by the carpenter, in order to obtain the maximum gross income? Formulate the above as a Linear Programming Problem and solve it, indicating clearly the feasible region in the graph.
The minimum value of z = 10x + 25y subject to 0 ≤ x ≤ 3, 0 ≤ y ≤ 3, x + y ≥ 5 is ______.
The minimum value of z = 2x + 9y subject to constraints x + y ≥ 1, 2x + 3y ≤ 6, x ≥ 0, y ≥ 0 is ______.
For the function z = 19x + 9y to be maximum under the constraints 2x + 3y ≤ 134, x + 5y ≤ 200, x ≥ 0, y ≥ 0; the values of x and y are ______.
For the LPP, maximize z = x + 4y subject to the constraints x + 2y ≤ 2, x + 2y ≥ 8, x, y ≥ 0 ______.
The maximum of z = 5x + 2y, subject to the constraints x + y ≤ 7, x + 2y ≤ 10, x, y ≥ 0 is ______.
Of all the points of the feasible region for maximum or minimum of objective function the points.
The feasible region (shaded) for a L.P.P is shown in the figure. The maximum Z = 5x + 7y is ____________.
Minimise z = – 3x + 4y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0 What will be the minimum value of z ?
Any point in the feasible region that gives the optional value (maximum or minimum) of the objective function is called:-
The shaded part of given figure indicates in feasible region, then the constraints are:
Solve the following linear programming problem graphically:
Minimize: Z = 5x + 10y
Subject to constraints:
x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x ≥ 0, y ≥ 0.
