English

A Small Firm Manufactures Necklaces and Bracelets. the Total Number of Necklaces and Bracelets that It Can Handle per Day is at Most 24 Formulate on L.P.P. for Finding How Many of Each Should Be Produced Daily to Maximize the Profit? - Mathematics

Advertisements
Advertisements

Question

A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is Rs 100 and that on a bracelet is Rs 300. Formulate on L.P.P. for finding how many of each should be produced daily to maximize the profit?

It is being given that at least one of each must be produced.

Solution

Let the number of necklaces manufacture be x,

and the number of bracelets manufacture be y.

since the total number of items are at most 24

`x + y <= 24`  .....(1)

Bracelets takes 1 hour to manufacture and necklaces takes half an hour to manufacture

x item takes x hour to manufacture and y items take y/2 hour to manufacture.

and maximum time available is 16 hours.

therefore

`x/2 + y <= 16` ....(2)

the profit on one necklace is Rs. 100 and the profit on one bracelet is Rs.300

Let the profit be Z. Now we wish to maximize the profit. So,

Max Z = 100x + 300y              ...(3)

So, `x + y <= 24`

`x/2 + y <= 16`

Max Z = 100x + 300y is the required L.P.P.

 

shaalaa.com
  Is there an error in this question or solution?
2016-2017 (March) Delhi Set 1

RELATED QUESTIONS

Solve the following Linear Programming Problems graphically:

Maximise Z = 5x + 3y

subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0


Solve the following Linear Programming Problems graphically:

Minimise Z = 3x + 5y

such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.


Solve the following Linear Programming Problems graphically:

Maximise Z = 3x + 2y

subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.


Show that the minimum of Z occurs at more than two points.

Minimise and Maximise Z = 5x + 10 y

subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0.


Show that the minimum of Z occurs at more than two points.

Minimise and Maximise Z = x + 2y 

subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.


Minimise Z = 13x – 15y subject to the constraints: x + y ≤ 7, 2x – 3y + 6 ≥ 0, x ≥ 0, y ≥ 0


Refer to Exercise 7 above. Find the maximum value of Z.


A manufacturer produces two Models of bikes-Model X and Model Y. Model X takes a 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hour available per week. Handling and Marketing costs are Rs 2000 and Rs 1000 per unit for Models X and Y respectively. The total funds available for these purposes are Rs 80,000 per week. Profits per unit for Models X and Y are Rs 1000 and Rs 500, respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.


A company makes 3 model of calculators: A, B and C at factory I and factory II. The company has orders for at least 6400 calculators of model A, 4000 calculator of model B and 4800 calculator of model C. At factory I, 50 calculators of model A, 50 of model B and 30 of model C are made every day; at factory II, 40 calculators of model A, 20 of model B and 40 of model C are made everyday. It costs Rs 12000 and Rs 15000 each day to operate factory I and II, respectively. Find the number of days each factory should operate to minimise the operating costs and still meet the demand.


The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y ______.

Compare the quantity in Column A and Column B

Column A Column B
Maximum of Z 325

The feasible solution for a LPP is shown in Figure. Let Z = 3x – 4y be the objective function. Minimum of Z occurs at ______.


Refer to Question 27. (Maximum value of Z + Minimum value of Z) is equal to ______.


Refer to Question 30. Minimum value of F is ______.


In a LPP, the objective function is always ______.


A corner point of a feasible region is a point in the region which is the ______ of two boundary lines.


If the feasible region for a LPP is unbounded, maximum or minimum of the objective function Z = ax + by may or may not exist.


Based on the given shaded region as the feasible region in the graph, at which point(s) is the objective function Z = 3x + 9y maximum?


A linear programming problem is as follows:

Minimize Z = 30x + 50y

Subject to the constraints: 3x + 5y ≥ 15, 2x + 3y ≤ 18, x ≥ 0, y ≥ 0

In the feasible region, the minimum value of Z occurs at:


For an objective function Z = ax + by, where a, b > 0; the corner points of the feasible region determined by a set of constraints (linear inequalities) are (0, 20), (10, 10), (30, 30) and (0, 40). The condition on a and b such that the maximum Z occurs at both the points (30, 30) and (0, 40) is:


In a linear programming problem, the constraints on the decision variables x and y are x − 3y ≥ 0, y ≥ 0, 0 ≤ x ≤ 3. The feasible region:


Z = 7x + y, subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0. The minimum value of Z occurs at ____________.


In linear programming infeasible solutions


In linear programming, optimal solution ____________.


In Corner point method for solving a linear programming problem, one finds the feasible region of the linear programming problem, determines its corner points, and evaluates the objective function Z = ax + by at each corner point. Let M and m respectively be the largest and smallest values at corner points. In case the feasible region is unbounded, m is the minimum value of the objective function.


Maximize Z = 3x + 5y, subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.


Maximize Z = 7x + 11y, subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×