Worksheet · Grade 12 · Applied Mathematics
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6 questions · 6 marks
In a linear programming problem, the quantities whose values are to be found, such as the number of units of each product to make, are called:
Answer: (b) decision variables
The decision variables (usually \(x\) and \(y\)) stand for the quantities the problem has to decide; the constraints and the objective function are then written in terms of them.
In a linear programming problem, the conditions \(x \ge 0\) and \(y \ge 0\) are called ______ constraints.
Answer: non-negative
They are the non-negative constraints: the decision variables stand for real quantities, such as units made or kilograms used, which cannot be negative.
The aim of every linear programming problem is to maximise something, such as profit.
Answer: False
Corrected statement: the aim of a linear programming problem is optimization, which may be to maximise (for example profit or revenue) or to minimise (for example cost or time) the objective function.
What is meant by the objective function of a linear programming problem?
Answer: It is the linear function of the decision variables, such as \(Z = ax + by\), whose value (profit, cost and so on) is to be maximised or minimised.
A bakery bakes \(x\) cakes and \(y\) loaves of bread a day. Its oven can bake at most 60 items a day. This condition is written as:
Answer: (c) \(x + y \le 60\)
The total number of items is \(x + y\), and 'at most 60' means it cannot go above 60, so \(x + y \le 60\). The option \(xy \le 60\) is not even linear.
Why does a business frame a linear programming problem? Answer in one sentence.
Answer: To find the best (optimal) way of using limited resources, such as money, machine hours, labour or raw material, so that a linear objective like profit is maximised or cost is minimised.
Build on it
8 questions · 11 marks
A tailor has 48 m of cloth. A kurta needs 2.5 m and a shirt needs 1.5 m of cloth. If she makes \(x\) kurtas and \(y\) shirts, the cloth constraint in its simplest form is:
Answer: (b) \(5x + 3y \le 96\)
Cloth used is \(2.5x + 1.5y \le 48\); multiplying both sides by 2 gives \(5x + 3y \le 96\). Doubling only the left side (giving 48) or swapping the coefficients are the traps.
Assertion (A): A farmer deciding how many hectares to sow with each of three crops needs three decision variables.
Reason (R): A linear programming problem has one decision variable for each quantity whose value has to be decided.
Answer: (a) Both A and R are true, and R is the correct explanation for A
The farmer must decide three quantities (the area under each crop), and R says each such quantity gets its own decision variable, so three variables are needed: R explains A.
A shop buys tea at ₹200 per kg and sells it at a profit of 20%; it buys coffee at ₹500 per kg and sells it at a profit of 15%. If it sells \(x\) kg of tea and \(y\) kg of coffee a week, the objective function (weekly profit, to be maximised) is \(Z =\) ______.
Answer: \(40x + 75y\)
Profit on 1 kg of tea is 20% of ₹200 = ₹40 and on 1 kg of coffee is 15% of ₹500 = ₹75, so the total profit is \(Z = 40x + 75y\).
Match each situation in Column A with the type of linear programming problem in Column B.
| Column A | Column B |
|---|---|
| (i) A hospital chooses how much of two foods to put in a patient's meal so that it has enough protein and iron at the least cost | (P) Transportation problem |
| (ii) A cement company decides how many tonnes to send from each of its two factories to three building sites so that the carriage cost is least | (Q) Allocation (investment) problem |
| (iii) A leather unit decides how many bags and belts to make each week with limited cutting and stitching hours, for the greatest profit | (R) Diet problem |
| (iv) A retired teacher decides how to split ₹6 lakh between a bank deposit and government bonds to earn the most yearly interest | (S) Manufacturing problem |
Answer: (i)-R, (ii)-P, (iii)-S, (iv)-Q
A least-cost food mix meeting nutrient needs is a diet problem; sending goods from sources to destinations at least cost is a transportation problem; deciding production with limited hours is a manufacturing problem; splitting a fixed sum among options is an allocation (investment) problem.
A factory makes \(x\) fans and \(y\) coolers a week. Using this factory as the example, explain the terms (a) decision variables, (b) constraint and (c) non-negative constraints of a linear programming problem.
Answer: (a) Decision variables are the quantities whose values are to be decided: here \(x\), the number of fans, and \(y\), the number of coolers made a week.
(b) A constraint is a linear inequality (or equation) in the decision variables that expresses a limit on a resource or a requirement: for example, if a fan needs 2 hours and a cooler 3 hours of assembly and only 60 hours are available, then \(2x + 3y \le 60\).
(c) Non-negative constraints are \(x \ge 0\) and \(y \ge 0\): the factory cannot make a negative number of fans or coolers.
How marks are given
A school hires \(x\) buses and \(y\) vans for a picnic. A bus carries 50 students and a van carries 12. At least 300 students must be carried. Write this condition as a linear inequality in its simplest form.
Answer: \(50x + 12y \ge 300\), i.e. \(25x + 6y \ge 150\).
A workshop makes \(x\) steel almirahs and \(y\) steel beds a week. An almirah needs 3 hours of welding and 2 hours of painting; a bed needs 2 hours of welding and 1 hour of painting. At most 60 hours of welding and 35 hours of painting are available, and at least 5 beds must be made for an order. Besides \(x \ge 0\), the constraints are:
Answer: (c) \(3x + 2y \le 60\), \(2x + y \le 35\), \(y \ge 5\)
Welding: \(3x + 2y \le 60\); painting: \(2x + y \le 35\) (both 'at most', so \(\le\)); the order needs 'at least 5 beds', so \(y \ge 5\).
Assertion (A): If a linear programming problem has the constraint \(x \ge 10\), then the constraint \(x \ge 0\) is automatically satisfied.
Reason (R): Non-negative constraints are never needed when a linear programming problem is written.
Answer: (c) A is true, but R is false
Any \(x \ge 10\) is also \(\ge 0\), so A is true. R is false: non-negative constraints are a standard part of every LPP, because decision variables stand for quantities that cannot be negative.
Stretch yourself
6 questions · 24 marks
A fruit seller sells \(x\) kg of mangoes and \(y\) kg of apples a day. She buys mangoes at ₹80 per kg and sells them at ₹100 per kg; she buys apples at ₹120 per kg and sells them at ₹150 per kg. She can spend at most ₹6,000 a day on buying fruit, and her cart holds at most 60 kg. To find the greatest daily profit, a student writes:
Maximise \(Z = 100x + 150y\) subject to \(80x + 120y \ge 6000\), \(x + y \le 60\).
Find the three mistakes and write the correct LPP.
Answer: Mistake 1: the objective uses selling prices, not profit. Profit is ₹20 per kg of mangoes and ₹30 per kg of apples, so \(Z = 20x + 30y\).
Mistake 2: the budget is 'at most ₹6,000', so the sign must be \(\le\): \(80x + 120y \le 6000\), i.e. \(2x + 3y \le 150\).
Mistake 3: the non-negative constraints \(x \ge 0\), \(y \ge 0\) are missing.
Correct LPP: Maximise \(Z = 20x + 30y\) subject to \(2x + 3y \le 150\), \(x + y \le 60\), \(x \ge 0\), \(y \ge 0\).
How marks are given
A company has two plants. Running plant P for an hour costs ₹2,000 and produces 6 units of item X and 3 units of item Y. Running plant Q for an hour costs ₹1,500 and produces 2 units of X and 4 units of Y. Each day at least 36 units of X and 30 units of Y are needed, and the two plants together can run for at most 12 hours. Formulate the LPP to find the least daily running cost.
Answer: Decision variables: let plant P run for \(x\) hours and plant Q for \(y\) hours a day (the decision is about hours, not units).
Objective: Minimise \(Z = 2000x + 1500y\).
Constraints:
Item X: \(6x + 2y \ge 36\), i.e. \(3x + y \ge 18\)
Item Y: \(3x + 4y \ge 30\)
Running time: \(x + y \le 12\)
\(x \ge 0\), \(y \ge 0\)
How marks are given
A poultry farm mixes two feeds, A and B. Each kg of feed A contains 300 g of protein and 100 g of fat; each kg of feed B contains 200 g of protein and 300 g of fat. Each day the flock needs at least 1.8 kg of protein and at least 1.2 kg of fat, and the feed hopper holds at most 10 kg. Feed A costs ₹30 per kg and feed B ₹25 per kg. Formulate the LPP to find the cheapest daily mix.
Answer: Let the mix contain \(x\) kg of feed A and \(y\) kg of feed B.
Minimise \(Z = 30x + 25y\).
Working in kg (300 g = 0.3 kg, 100 g = 0.1 kg, 200 g = 0.2 kg, 300 g = 0.3 kg):
Protein: \(0.3x + 0.2y \ge 1.8\), i.e. \(3x + 2y \ge 18\)
Fat: \(0.1x + 0.3y \ge 1.2\), i.e. \(x + 3y \ge 12\)
Hopper: \(x + y \le 10\)
\(x \ge 0\), \(y \ge 0\)
How marks are given
A handloom unit weaves sarees and dupattas. A saree takes 4 hours 30 minutes on the loom and a dupatta 1 hour 30 minutes, and the loom can run for at most 45 hours a week. A saree sells for ₹2,400 and uses yarn and dye costing ₹1,500; a dupatta sells for ₹800 and uses yarn and dye costing ₹450. The weaver is paid fixed wages of ₹6,000 a week, whatever is woven. A shop has ordered at least 6 dupattas, and the unit cannot sell more than two dupattas for every saree it weaves. (a) Find the profit on each saree and on each dupatta before wages. (b) Formulate the LPP to find the greatest weekly profit after wages, writing each constraint in its simplest form. (c) Explain why the fixed wages do not change which weekly plan is best.
Answer: (a) Profit on a saree = ₹2,400 - ₹1,500 = ₹900; profit on a dupatta = ₹800 - ₹450 = ₹350.
(b) Decision variables: let the unit weave \(x\) sarees and \(y\) dupattas a week.
Objective: Maximise \(Z = 900x + 350y - 6000\).
Loom: 4 hours 30 minutes = 4.5 hours and 1 hour 30 minutes = 1.5 hours, so \(4.5x + 1.5y \le 45\), i.e. \(3x + y \le 30\).
Order: \(y \ge 6\).
Sales: at most two dupattas for every saree, so \(y \le 2x\), i.e. \(2x - y \ge 0\).
Non-negative constraints: \(x \ge 0\), \(y \ge 0\).
(c) The ₹6,000 is the same for every plan, so it lowers the profit of every plan by the same amount. The plan that makes \(900x + 350y\) greatest also makes \(Z\) greatest; the wages change only the value of the greatest profit, not the plan.
How marks are given
A trader has two warehouses: A holds 70 tonnes of wheat and B holds 50 tonnes. Market M1 needs 60 tonnes and market M2 needs 40 tonnes. The cost of carrying one tonne is:
A to M1: ₹50
A to M2: ₹40
B to M1: ₹30
B to M2: ₹60
Let \(x\) tonnes go from A to M1 and \(y\) tonnes from A to M2, with B sending the rest of each market's need. Formulate the LPP to find the least total carrying cost, writing the objective function in its simplest form.
Answer: B sends \(60 - x\) tonnes to M1 and \(40 - y\) tonnes to M2.
Total cost:
\(Z = 50x + 40y + 30(60 - x) + 60(40 - y)\)
\(= 50x + 40y + 1800 - 30x + 2400 - 60y\)
\(= 20x - 20y + 4200\)
Minimise \(Z = 20x - 20y + 4200\) subject to:
Stock at A: \(x + y \le 70\)
Stock at B: \((60 - x) + (40 - y) \le 50\), i.e. \(x + y \ge 50\)
No amount negative: \(x \ge 0\), \(y \ge 0\), \(60 - x \ge 0\) (so \(x \le 60\)), \(40 - y \ge 0\) (so \(y \le 40\)).
How marks are given
A juice company in Jaipur is planning its advertising for May. It can advertise on a local TV channel and on a local FM radio station. One TV advertisement costs ₹50,000 and is seen by about 2,00,000 people; one radio advertisement costs ₹10,000 and is heard by about 30,000 people. The company's advertising budget for the month is ₹5,00,000. The TV channel has only 8 advertisement slots free in May, so the company cannot book more than 8 TV advertisements. The marketing head also wants at least 5 radio advertisements, because many customers listen to the radio while driving to work. The company wants to choose the number of TV and radio advertisements so that the total number of people reached is as large as possible. A trainee in the marketing team is asked to write this as a linear programming problem before anyone solves it.
(i) \(x\) = number of TV advertisements and \(y\) = number of radio advertisements booked in May.
(ii) Maximise \(Z = 200000x + 30000y\), the total number of people reached.
(iii) \(50000x + 10000y \le 500000\), i.e. \(5x + y \le 50\).
(iv) Other constraints: \(x \le 8\), \(y \ge 5\), \(x \ge 0\), \(y \ge 0\).
For \(x = 8\), \(y = 12\): \(x \le 8\) and \(y \ge 5\) hold, but \(5(8) + 12 = 52 > 50\): the plan costs ₹4,00,000 + ₹1,20,000 = ₹5,20,000, more than the ₹5,00,000 budget. So the plan is not allowed.
