Advertisements
Advertisements
Question
A furniture manufacturing company plans to make two products : chairs and tables. From its available resources which consists of 400 square feet to teak wood and 450 man hours. It is known that to make a chair requires 5 square feet of wood and 10 man-hours and yields a profit of Rs 45, while each table uses 20 square feet of wood and 25 man-hours and yields a profit of Rs 80. How many items of each product should be produced by the company so that the profit is maximum?
Solution
Let x units of chairs and y units of tables were produced Therefore \[x, y \geq 0\]
The given information can be tabulated as follows:
Wood(square feet) | Man hours | |
Chairs(x) | 5 | 10 |
Tables(y) | 20 | 25 |
Availability | 400 | 450 |
Therefore, the constraints are
\[5x + 20y \leq 400\]
\[10x + 25y \leq 450\]
It is known that to make a chair requires 5 square feet of wood and 10 man-hours and yields a profit of Rs 45, while each table uses 20 square feet of wood and 25 man-hours and yields a profit of Rs 80.
Therefore, profit gained to make x chairs and y tables is Rs 45x and Rs 80y respectively.
Total profit = Z = \[45x + 80y\]
which is to be maximised.
Max Z = \[45x + 80y\]
subject to
\[5x + 20y \leq 400\]
\[10x + 25y \leq 450\]
First we will convert inequations into equations as follows:
5x + 20y = 400, 10x + 25y = 450, x = 0 and y = 0
Region represented by 5x + 20y ≤ 400:
The line 5x + 20y = 400 meets the coordinate axes at
5x + 20y = 400 . Clearly (0,0) satisfies the 5x + 20y = 400 . So, the region which contains the origin represents the solution set of the inequation 5x + 20y ≤ 400.
Region represented by 10x + 25y ≤ 450:
The line 10x + 25y = 450 meets the coordinate axes at \[C\left( 45, 0 \right)\] and \[D\left( 0, 18 \right)\] respectively. By joining these points we obtain the line
10x + 25y = 450. Clearly (0,0) satisfies the inequation 10x + 25y ≤ 450. So,the region which contains the origin represents the solution set of the inequation 10x + 25y ≤ 450.
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 5x + 20y ≤ 400, 10x + 25y ≤ 450, x ≥ 0, and y ≥ 0 are as follows.

The corner points are A(0, 18), B(45, 0)
The values of Z at these corner points are as follows
Corner point | Z= 45x + 80y |
A | 1440 |
B | 2025 |
The maximum value of Z is 2025 which is attained at B \[\left( 45, 0 \right)\] .
Thus, the maximum profit is of Rs 2025 obtained when 45 units of chairs and no units of tables are produced
APPEARS IN
RELATED QUESTIONS
Minimize : Z = 6x + 4y
Subject to the conditions:
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 retired person wants to invest an amount of Rs. 50, 000. His broker recommends investing in two type of bonds ‘A’ and ‘B’ yielding 10% and 9% return respectively on the invested amount. He decides to invest at least Rs. 20,000 in bond ‘A’ and at least Rs. 10,000 in bond ‘B’. He also wants to invest at least as much in bond ‘A’ as in bond ‘B’. Solve this linear programming problem graphically to maximise his returns.
Solve the following L.P.P. graphically:
Minimise Z = 5x + 10y
Subject to x + 2y ≤ 120
Constraints x + y ≥ 60
x – 2y ≥ 0 and x, y ≥ 0
Maximize Z = 15x + 10y
Subject to
\[3x + 2y \leq 80\]
\[2x + 3y \leq 70\]
\[ x, y \geq 0\]
Maximize Z = 4x + 3y
Subject to
\[3x + 4y \leq 24\]
\[8x + 6y \leq 48\]
\[ x \leq 5\]
\[ y \leq 6\]
\[ x, y \geq 0\]
Minimize Z = x − 5y + 20
Subject to
\[x - y \geq 0\]
\[ - x + 2y \geq 2\]
\[ x \geq 3\]
\[ y \leq 4\]
\[ x, y \geq 0\]
Maximize Z = 2x + 3y
Subject to
\[x + y \geq 1\]
\[10x + y \geq 5\]
\[x + 10y \geq 1\]
\[ x, y \geq 0\]
Solved the following linear programming problem graphically:
Maximize Z = 60x + 15y
Subject to constraints
\[x + y \leq 50\]
\[3x + y \leq 90\]
\[ x, y \geq 0\]
A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 of calories. Two foods A and B, are available at a cost of Rs 4 and Rs 3 per unit respectively. If one unit of A contains 200 units of vitamin, 1 unit of mineral and 40 calories and one unit of food B contains 100 units of vitamin, 2 units of minerals and 40 calories, find what combination of foods should be used to have the least cost?
To maintain one's health, a person must fulfil certain minimum daily requirements for the following three nutrients: calcium, protein and calories. The diet consists of only items I and II whose prices and nutrient contents are shown below:
Food I | Food II | Minimum daily requirement | |
Calcium Protein Calories |
10 5 2 |
4 6 6 |
20 20 12 |
Price | Rs 0.60 per unit | Rs 1.00 per unit |
Find the combination of food items so that the cost may be minimum.
A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A in the diet? What is the minimum of vitamin A.
A manufacturer has three machines installed in his factory. machines I and II are capable of being operated for at most 12 hours whereas Machine III must operate at least for 5 hours a day. He produces only two items, each requiring the use of three machines. The number of hours required for producing one unit each of the items on the three machines is given in the following table:
Item | Number of hours required by the machine | ||
A B |
I | II | III |
1 2 |
2 1 |
1 5/4 |
He makes a profit of Rs 6.00 on item A and Rs 4.00 on item B. Assuming that he can sell all that he produces, how many of each item should he produces so as to maximize his profit? Determine his maximum profit. Formulate this LPP mathematically and then solve it.
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 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?
A firm makes items A and B and the total number of items it can make in a day is 24. It takes one hour to make an item of A and half an hour to make an item of B. The maximum time available per day is 16 hours. The profit on an item of A is Rs 300 and on one item of B is Rs 160. How many items of each type should be produced to maximize the profit? Solve the problem graphically.
A manufacturer has three machine I, II, III installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must be operated for atleast 5 hours a day. She produces only two items M and N each requiring the use of all the three machines.
The number of hours required for producing 1 unit each of M and N on the three machines are given in the following table:
Items | Number of hours required on machines | ||
I | II | III | |
M | 1 | 2 | 1 |
N | 2 | 1 | 1.25 |
She makes a profit of ₹600 and ₹400 on items M and N respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit?
The value of objective function is maximum under linear constraints ______.
Maximize: z = 3x + 5y Subject to
x +4y ≤ 24 3x + y ≤ 21
x + y ≤ 9 x ≥ 0 , y ≥0
A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A
require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours and 20 minutes available for cutting and 4 hours available for assembling. The profit is Rs. 50 each for type A and Rs. 60 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize profit? Formulate the above LPP and solve it graphically and also find the maximum profit.
A manufacturer has employed 5 skilled men and 10 semi-skilled men and makes two models A and B of an article. The making of one item of model A requires 2 hours of work by a skilled man and 2 hours work by a semi-skilled man. One item of model B requires 1 hour by a skilled man and 3 hours by a semi-skilled man. No man is expected to work more than 8 hours per day. The manufacturer's profit on an item of model A is ₹ 15 and on an item of model B is ₹ 10. How many items of each model should be made per day in order to maximize daily profit? Formulate the above LPP and solve it graphically and find the maximum profit.
The graph of the inequality 3X − 4Y ≤ 12, X ≤ 1, X ≥ 0, Y ≥ 0 lies in fully in
Find the feasible solution of linear inequation 2x + 3y ≤ 12, 2x + y ≤ 8, x ≥ 0, y ≥ 0 by graphically
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 ______.
The feasible region of an LPP is shown in the figure. If z = 3x + 9y, then the minimum value of z occurs at ______.
The constraints of an LPP are 7 ≤ x ≤ 12, 8 ≤ y ≤ 13. Determine the vertices of the feasible region formed by them.
Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z = px + qy, where p, q > 0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is ______.
Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded, then ____________.
The feasible region (shaded) for a L.P.P is shown in the figure. The maximum Z = 5x + 7y is ____________.
Which of the statements describe the solution set for `-2(x + 8) = - 2x + 20`?
The maximum value of 2x + y subject to 3x + 5y ≤ 26 and 5x + 3y ≤ 30, x ≥ 0, y ≥ 0 is ______.
The shaded part of given figure indicates in feasible region, then the constraints are:
The objective function Z = x1 + x2, subject to the constraints are x1 + x2 ≤ 10, – 2x1 + 3x2 ≤ 15, x1 ≤ 6, x1, x2 ≥ 0, has maximum value ______ of the feasible region.
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.
The feasible region corresponding to the linear constraints of a Linear Programming Problem is given below.
Which of the following is not a constraint to the given Linear Programming Problem?
If x – y ≥ 8, x ≥ 3, y ≥ 3, x ≥ 0, y ≥ 0 then find the coordinates of the corner points of the feasible region.