scholarship questions

Scholarship Questions

2003

2003

Question 4• An oil company operates two refineries. The refineries produce

three types of fuel – aviation grade, regular grade petrol and super grade petrol.

• • Refinery 1 costs $160 000 per day to operate and Refinery 2

costs $175 000 per day to operate.• • The oil company has contracts to produce at least 120 000

litres of aviation fuel, 300 000 litres of regular grade petrol and 108 000 litres of super grade petrol per month.

• • The following table gives the daily production statistics in litres:

• Let the number of days per month that Refinery 1 operates be x and the number of days per month that Refinery 2 operates be y.

• • (a) Using graph paper where necessary,

determine the optimal number of days that each refinery should operate in order to minimise the total cost of operating them both. Calculate this minimum cost.

• Constraints:• 10 000x + 10 000y ≥ 120 000 Aviation• 20 000x + 30 000y ≥ 300 000 Regular• 6 000x + 18 000y ≥ 108 000 Super• 0 ≤ x ≤ 31 Restriction on days• 0 ≤ y ≤ 31

Simplify the inequalities:

• Constraints:• x + y ≥ 12 Aviation• 2x + 3y ≥ 30 Regular• x + 3 y ≥ 18 Super• 0 ≤ x ≤ 31 Restriction on days• 0 ≤ y ≤ 31

• Objective function is• C = 160 000x + 175 000y

• Gradient: -160/175 = -32/35

• Difficult to use this so look for intersection points.

(0,12), (6,6), (12,2), (18,0),

Costs at each vertex ($000)

Optimum number of days for both refineries is 6Minimum cost = $2 010 000

• The company is investigating possible changes at Refinery 2 that may alter the daily cost of operating the refinery. Assuming that the daily cost of operating Refinery 1 remains fixed at $160 000, determine the range of values that the daily cost of operating Refinery 2 could lie within and still give the production combination in (a).

The gradient needs to stay between the gradients of the 2 lines that form the intersection (6,6) i.e. between -1 and -2/3

2004

Question 4

Question 4

• To produce one bottle of POW takes 30 minutes and to produce one bottle of ZAP takes 20 minutes, and there are 25 hours available in total for this production-run operation.

• x = bottles of POW• y = bottles of ZAP

Question 4

• To produce one bottle of POW takes 30 minutes and to produce one bottle of ZAP takes 20 minutes, and there are 25 hours available in total for this production-run operation.

Question 4

• Availability of a particular additive means that the chain cannot produce more than 35 bottles of POW and a combined total of 65 bottles on any production run.

Question 4

• Availability of a particular additive means that the chain cannot produce more than 35 bottles of POW and a combined total of 65 bottles on any production run.

Question 4

• As a minimum production requirement, at least 15 bottles of each weed killer must be produced in each run.

Question 4a

• The management of NAILS wants to know how many bottles of both POW and ZAP should be produced. The preliminary estimates of their potential profitability are $20 per bottle of POW and $10 per bottle of ZAP. Perform an appropriate analysis and make a recommendation to management about the amount of POW and ZAP that could be produced to maximise profit.

Question 4

• Profit

• Gradient is -2

Feasible region

(15, 15) (35, 15), (35, 22.5), (20, 45), (15, 50)

Feasible region

Blowup of the section

Feasible region

(35, 22.5) = (35, 22) and (34, 24) as need integer values

Question 4b

• Suppose that the profit of $20 per bottle of POW was overestimated and it was in reality only $15 per bottle. How does that change your production recommendation in part (a)?

Profit = 15x + 10y

Profit = 15x + 10y

• This has the same gradient of our first constraint

• Can have any of (20, 45), (22, 42), (24, 39), (26, 36), (28, 33), (30, 30), (32, 27), (34, 24)

• It is found that the sales of POW and ZAP are closely related by the function y = 9 ln(x),

• where x = number of bottles of POW and y = number of bottles of ZAP.

• Management therefore requests that the ratio “number of bottles of POW produced : number of bottles of ZAP produced be x : 9 ln(x)”. The constraints need to be satisfied and the estimated profit needs to be maximised, based on a profit of $15 per bottle of POW and $10 per bottle of ZAP, so as to mirror demand.

y = 9lnx

Equation

Solve the equation derived in part (c) and make a recommendation to management about the amount

of POW and ZAP that could be produced to maximise profit.

Solve the equation derived in part (c) and make a recommendation to management about the amount

of POW and ZAP that could be produced to maximise profit.

• Use your graphics calculator:• This gives 29.661…• Integer values• 29 bottles POW and 31 bottles ZAP

2005

Question 1

Question 1

• The refreshment bar sells two types of pie, steak and mince. The number of pies ordered each day needs

• to satisfy the following daily constraints.

At least 42 pies will be sold in total

A minimum of 24 mince pies will be sold

No more than 3 mince pies will be sold for every steak pie sold

At most, 12 more steak pies will be sold than mince pies.

Feasible region

• The refreshment bar pays $1.30 for each steak pie and $1.00 for each mince pie.

Objective function

Minimum cost

• (24, 36) next point will be less• (24, 18) $47.40• (10.5, 31.5) $45.15• Cannot buy half a pie

Minimum cost

• Investigate to get minimum at• (31, 11)• 31 mince and 11 steak pies

• Assume the cost of steak pies remains at $1.30 each. The cost of mince pies increases so that the optimal solution is no longer that obtained in (a). State the possible changes in the cost of mince pies that give rise to these different optimal solution(s), and give the new optimal solution(s) for each of these changes.

Feasible region

Question 1b

• The price would have to be the same as the steak pie, I.e. $1.30 as the gradient of the line is -1. Then all integer points between x= 24 and 31 would give the same minimum

• I.e. (24,18), (25, 17), (26, 16), (27, 15), (28, 14), (29, 13), (30, 12) and our result (31, 11). Once the cost was greater than $1.30, the minimum is achieved at (24, 18)

• Suppose the demand for the pies is modelled by the function:

• S 2 = 66M + 10 where • S represents the daily demand for steak

pies,• and M represents the daily demand for

mince pies.

1c Find the optimal solution that satisfies the daily demand

Demand enters the feasible Region here

Minimum value- check integer values

Solve for the intersection

Solve for the intersection

Integer answer is 39, giving y = 51

This satisfies all constraints and minimises the cost.

2006

Note: before you even start

• All questions are based on a product, strawberry yoghurt, produced by a company named Fastidious Foods. This company produces two types of strawberry yoghurt, light and standard, in  150 mL pottles. Each type of yoghurt is packaged in boxes containing six pottles.

Question 1

• The number of boxes of each type produced per day needs to satisfy the following constraints:

The total number of boxes produced is not to exceed 400

And there must be at least 200 boxes of standard yoghurt produced.

The number of boxes of standard yoghurt produced is to be at least twice the number of

light yoghurt produced.

No more than seven boxes of standard yoghurt are to be produced for every two boxes of light yoghurt produced.

Feasible region

A box of light yoghurt yields a greater profit than a box of standard yoghurt. 

Gradient < -1 (i.e. steeper line) makes green/red intersection maximum i.e. (133, 267)

Constraints 2 and 3 in (a) are changed so that the number of boxes of light yoghurt produced is between 33.3% and 40.0% of the number of boxes of standard yoghurt produced. Assume that constraint 1 in (a) still applies and that each box of light yoghurt still yields a

greater profit than a box of standard yoghurt.   

Find the number of boxes of each type of yoghurt that should be produced now to maximise the

profit.

1b. New constraints

Optimum solution at intersection of red/green lines as gradient < -1 ( i.e. steeper line)

Solution is (114, 286); 114 light, 286 standard

2007

Question 1

• Shedz processes two varieties of kiwifruit – green and gold. Constraints on daily packing are as follows:

Question 1

• Let x = number of trays of green• and y = number of trays of gold

At most, 2 000 trays can be packed in total.

• x + y ≤ 2000

At most, 1 200 trays of green can be packed.

• x ≤ 1200

A minimum of 200 trays of gold need to be packed.

• y ≥ 200

The ratio of the number of packed trays of green to gold

must be no more than 5:2.

• 2x ≤ 5y

Constraints

• x + y ≤ 2000• x ≤ 1200• y ≥ 200• 2x ≤ 5y

Feasible region

The ratio of the profit per tray of the green variety to

the gold is 12:13.• Objective function is:

Profit = 12mx + 13my where m is a multiple constant

Profit = 12mx + 13my

• Optimum point is x = 0 and y = 2000• So 0 trays of green and 2000 trays of gold

should be packed daily to maximise the profit.

The output of Shedz should reflect market demand and maintain a strong profit.

• Given that there is a market demand for both green and gold varieties, suggest what other solutions to part (a) (i) are possible that do not involve changing any constraints.

• Justify your suggestions fully.

1 (a) (ii)1. It’s highly unlikely that all 2000 gold would be

sold. We know that we will sell some green due to demand. In fact if x = 600 and y = 1400, we would not have to sell about 50 gold for this to be a more profitable combination.

2. Objective function is almost parallel to constant boundary line x + y = 2000. Small differences of x and y from the optimal value don’t affect the profit significantly.

Feasible region

2008

Question 6

• A new hotel for Statsmod Enterprises is to have only deluxe and standard rooms. In designing the new hotel, the following constraints need to apply:

Define your variables

• Let x = number of standard rooms• Let y = number of deluxe rooms

Constraints

• The maximum number of rooms is 65.

• The building cost budget for the rooms has a maximum of $3 million. Building costs are $45 000 for a standard room and $60 000 for a deluxe room.

• When the number of standard rooms (x) is such that 10m < x ≤ 10(m + 1), then the number of deluxe rooms (y) is such that y ≤ 5(m + 1), where m = 0, 1, 2, 3, 4.

This is a step function

• When the number of standard rooms (x) is such that 10m < x ≤ 10(m + 1), then the number of deluxe rooms (y) is such that y ≤ 5(m + 1), where m = 0, 1, 2, 3, 4.

This is a step function

m x0 0<x≤10 Y≤51 10<x≤20 Y≤102 20<x≤30 Y≤153 30<x≤40 y≤204 40<x≤50 Y≤25

Feasible region

• The profit for deluxe rooms is 40% greater than that for standard rooms.

• Assuming that all rooms are occupied, find the number of each type of room that should be built to ensure maximum profit.A grid is provided on page 26 of the Answer Booklet to help you answer this question.

Maximum occurs (40,20)

(b)

• Statsmod Enterprises wants to ensure maximum profit and build the maximum number of 65 rooms, but cannot do so within the $3 million building cost budget.

• What is the smallest required increase in this budget?

We are looking at the blue line and then calculating profit

X Y Profit41 24 74.642 23 74.243 22 73.844 21 73.445 20 73

Max (41, 24)

X Y Profit41 24 74.642 23 74.243 22 73.844 21 73.445 20 73

Max (41, 24)

• Cost:

• Increase = $285,000

• Need minimum cost to produce maximum profit.

(c)

• Suppose the mean occupancy rate for deluxe rooms is 60% and that all three constraints still apply.

• While still ensuring maximum profit, calculate the occupancy rate for standard rooms, which will give more than one solution for each type of room that should be built. Clearly state all of these solutions.

• Suppose the mean occupancy rate for deluxe rooms is 60% and that all three constraints still apply.

• While still ensuring maximum profit, calculate the occupancy rate for standard rooms, which will give more than one solution for each type of room that should be built. Clearly state all of these solutions.

2009

Question 6

• A proposed housing development will have two types of home, single detached homes and duplex units (a block of two homes). Two of the resources available are six hectares (60 000 m2) of land for housing and a building budget of $7 million.

• A single detached home requires 600 m2 of land and costs $65 000 to build. A duplex unit requires 800 m2 of land and costs $100 000 to build.

• The profit from the sale of a duplex unit is 50% more than that from a single detached home.

• Let • x be the number of single detached homes

and • y be the number of duplex units. Assume

that all homes built are sold.

A single detached home requires 600 m2 of land and costs $65 000 to build. A duplex unit requires 800 m2 of land and costs

$100 000 to build.

• Let x = number of single detached homes• Let y = number of duplex units• The constraints are:• Land: 600x + 800y ≤ 60 000 • simplified 3x + 4y ≤ 300• Costs: 65x + 100y ≤ 7 000 • simplified 13x + 20y ≤ 1400• Non-negativity: x ≥ 0, y ≥ 0

Find the number of each type of home that should be built to maximise the profit. A grid is provided on page 26 of the

Answer Booklet to help you answer this question.

• The profit function is • P = k(x + 1.5y)

Feasible region

50, 37.5 need to look at points

Feasible region

Gradient of objective function is -2/3

Start at (40, 44) and go down 2 in y and up 3 in x until (52, 36)

To ensure maximum profit ( P = 106k in each case)

• x = 40 and y = 44 that is 40 single detached homes and 44 duplex units should be built or

• x = 43 and y = 42 that is 43 single detached homes and 42 duplex units should be built or

• x = 46 and y = 40 that is 46 single detached homes and 40 duplex units should be built or

• x = 49 and y = 38 that is 49 single detached homes and 38 duplex units should be built or

• x = 52 and y = 36 that is 52 single detached homes and 36 duplex units should be built.

Judgement

• S: Any one of these optimal point(s). • P: All two constraints correct OR any

constraint correct plus the correct profit objective function

• P: Feasible region clearly shown on graph as opposed to constraint equations.

Note:

• 1. Can use graph as evidence for the constraints.

• 2. Profit function can take the form • P = 1.0x + 1.5y.

Question 6b

• A new council regulation requires that there must be at least three single detached homes for every two duplex units in the new development.

• Calculate the percentage reduction in maximum profit resulting from this regulation.

• New Council regulation: 3y ≤ 2x

Now to ensure maximum profit, x = 53 and y = 35 or x = 56 and y = 33.

Feasible region

• For either optimal point, % reduction in maximum profit =

• = 0.47%.

Judgement

• S: Correct method and answer or correct new optimal point and carried error in % reduction in profit due to wrong answer in (a). new optimal point.

• P: Identification of correct new optimal point

• N: Incorrect

Note:

• 1. Only one optimal point is required.

• From previous similar developments, it has been shown that, for social reasons, it is desirable that the relationship between the numbers of the two types of homes built is approximately modelled by y = 35e-0.014x. Taking this relationship into account and the constraint in (b), calculate the amount of unused land and / or unused budget when x and y are such that the profit is maximised.

With social constraint, point where profit is maximised is given by: x = 85 and y = 11 with profit = $101.5k

Intersection point on calculator is 86, 10.5

• Unused land = 60 000 - 800(11) - 600(85) = 200 m2

• Unused budget = 7 000 - 100(11) - 65(85) = $375 000

2010

Question 2b

• Two different models of Statsmobiles are manufactured, X and Y.

• Let x be the number of model X manufactured annually and y be the number of model Y manufactured annually.

• Annual demand for Statsmobiles is at least • 6 000.

• Annual demand for model Y is between 95% and 170% of the annual demand for model X.

•

In each of the following three cases, find the optimal solution(s) that minimises the annual manufacturing cost, while meeting the above demand constraints.

The cost of producing each model Y is less than the cost of producing each model X.

Feasible region

The cost of producing each model Y is less than the cost of producing each model X

Feasible region

The cost of producing each model Y is less than the cost of producing each model X

Feasible region

The cost of producing each model Y is less than the cost of producing each model X

Feasible region

When A<1, gradient is steeper and so red/blue intersection is a minimum i.e. (x = 2 223 and y = 3 777)

The cost of producing each model Y is equal to the cost of producing each model X.

Feasible region

When A=1, gradient is the same as the blue line i.e. where 2 223 ≤ x ≤ 3 076 )

ThThe cost of producing each model Y is greater than the cost of producing each model X.

Feasible region

When A>1, gradient is less than the blue line i.e. x = 3 076 and y = 2 924

2011

Question 5

• This supplier operates out of two warehouses: eastern and western. The supplier receives orders from two customers, A and B, who require sheets of plywood.

Delivery costs per sheet are:

• $0.50 from the eastern warehouse to customer A

• $K (K is a constant) from the eastern warehouse to customer B

• $0.40 from the western warehouse to customer A

• $0.55 from the western warehouse to customer B.

• Let x represent the number of sheets delivered from the eastern warehouse to customer A and

• y represent the number of sheets delivered from the eastern warehouse to customer B.

• Customer A needs 50 sheets and customer B needs 70 sheets.

Customer A needs 50 sheets and customer B needs 70 sheets.

Customer A needs 50 sheets and customer B needs 70 sheets.

A BEastern x yWestern 50 - x 70 - y

• The eastern warehouse has 80 sheets in stock and

• the western warehouse has 45 sheets in stock.

• Use K = 0.60. • Identify all the constraints in terms of x and y. • Draw the feasible region on the graph paper

provided on page 26 of the Answer Booklet. • Write the overall delivery cost in terms of x and y. • Describe the delivery plan from each warehouse

to each customer that minimises the overall delivery cost.

Add non-negativity

Feasible region is between red and green lines

Objective Function

Gradient of objective function is -2

Minimum at (5, 70)

Delivery plan

• Customer A needs 50 sheets• 5 sheets from eastern warehouse• 45 sheets from western warehouse

• Customer B needs 70 sheets• All 70 from eastern warehouse

(ii)

• Find the value of K that gives multiple delivery plans and also minimises the overall delivery cost. State this minimum overall delivery cost.

(ii)

• Find the value of K that gives multiple delivery plans and also minimises the overall delivery cost. State this minimum overall delivery cost.

• We need the gradient to be the same as constraint

• i.e. -1

• Substitute to find the minimum cost:• Use any integer point on the line (5, 70)

• O + S: Delivery Plan and minimum overall delivery cost correct.

• S: Either (i) or (ii) correct.2P: Both constraints correct in (i) and K correct in (ii).P: Constraints correct in (i) or K correct in (ii).

(b)

• This part investigates the effect of changes made to the situation in part (a). Customer B still needs 70 sheets and the eastern warehouse still has 80 sheets in stock.

• For each of the following find the values of x and y that minimise the overall delivery cost.

• Use K = 0.60, customer A still needs 50 sheets, and the western warehouse has 40 sheets in

• stock (instead of 45).

• Use K = 0.60, customer A still needs 50 sheets, and the western warehouse has 40 sheets in

• stock (instead of 45).

Feasible region is now the pink line 10 < x < 50

Point (10, 70) to get a cost of $66.50

Constraint changes

• Use K = 0.70, customer A needs 48 sheets (instead of 50), and the western warehouse has 45 sheets in stock.

• Use K = 0.70, customer A needs 48 sheets (instead of 50), and the western warehouse has 45 sheets in stock.

Minimum at (48, 25)

2012

The gym is planning on purchasing some new treadmills, steppers and rowing machines. In order to meet usage requirements, there are to be a minimum of eight treadmills, five steppers and three rowing machines. The floor area needed for each piece of equipment is 3 m2 per treadmill, 2 m2 per stepper and 4 m2 per rowing machine. After allowing for room between each piece of equipment, 132 m2 is available. All of this area must be used. Data collected on concurrent usages in peak periods determined that there should be no more than three steppers to every two treadmills and at least one stepper to every treadmill.

In order to meet usage requirements, there are to be a minimum of eight treadmills, five steppers and three rowing machines.

The floor area needed for each piece of equipment is 3 m2 per treadmill, 2 m2 per stepper and 4 m2 per rowing machine. After allowing for room between each piece of equipment, 132 m2 is available. All of this area must be used.

Data collected on concurrent usages in peak periods determined that there should be no more than three steppers to every two treadmills and at least one stepper to every treadmill.

Data collected on concurrent usages in peak periods determined that there should be no more than three steppers to every two treadmills and at least one stepper to every treadmill.

Objective function

The cost of each piece of equipment is in the ratio 2:3:5 for treadmills, steppers and rowing machines respectively.

Let x be the number of treadmills and y be the number of steppers.

(a) Show that 3x +2y ≤ 120

Let z be the number of rowing machines

(b)

Find the number of treadmills, steppers and rowing machines that should be purchased and installed in order to minimise the cost.

GRAPH

Feasible region

Vertices: (24, 24), (20, 30), (8,8), (8,12)

(24, 24)

135k

(20, 30) 145k(8,8) 155k

(8,12) 157

Minimum cost for x = 24, y = 24

Optimal point (x , y, z ) = (24, 24, 3) which represents 24 treadmills, 24 steppers and 3 rowing machines.

(c)

Suppose that the cost of each piece of equipment had been in the ratio 2:3:c for treadmills, steppers and rowing machines respectively. Given that all the other constraints still apply and that the cost is to be minimised, find the value of c that gives multiple solutions.

New cost function

Multiple solutions occur when the cost function is parallel to one of the sides of the feasible region and the cost is minimised.Gradient of cost function