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Leaving Certificate Mathematics Ordinary Level 2000

Question 11

(a) The line K passes through the points (2, 0) and(0, 4).

(i) Find the equation of the line K.

(ii) Write down three inequalities whichdefine the shaded region in the diagram.

(b) Two types of machines, type A and type B, can be purchased for anew factory. Each machine of type A costs IR1600. Eachmachine of type B costs IR800. The purchase of the machinescan cost, at most, IR27 200.

Each machine of type A needs 90m2

of floor space in the factory.

Each machine of type B needs 54m2 of floor space.

The maximum amount of floor space available for the machines is 1620m2.

(i) If x represents the number of machines of type A and yrepresents the number of machines of type B, write down two

inequalities in x and y and illustrate these on graph paper.

(ii) The daily income from the use of each machine of type A isIR75. The daily income from the use of each machine oftype B machine is IR42. How many of each type ofmachine should be purchased so as to maximise dailyincome?

(iii) What is the maximum daily income?

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Answer

(a)(i) Using the coordinate Geometry formula

y-y1 = m(x x1) we can find the equation of the line. Firstly we need to findmthe slope using the formula

y2-y1______

x2-x1

Let (x1,y1) be (2,0) and (x2,y2) be equal to (0,4). Then we get 4 0______

0 2

which equals -2. The slope mis now equal to 2.

Then the equation of the line using the point (2,0) and m= -2 is

y 0= -2(x 2)

y = -2x + 4

2x + y 4 = 0 is the equation of K.

(ii) the three inequalities are

1. the X axis Y0

2. the Y axis X03. 2x + y 4 = 0

(b) (i)

1600x + 800y 27,200 cancel 2 zeroes

90x + 54y 1620 divide across by 9

We now have 10x + 6y 180

5x + 3y 90

10x + 6y 180 if x = 0 y = 34 then we have the coordinates (0,34)

if y = 0 x = 17 then we have the coordinates (17,0)

5x + 3y 90 if x = 0 y = 30 then we have the coordinates (0,30)

if y = 0 x = 18 then we have the coordinates (18,0)

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To find their point of Intersection we solve them simultaneously as follows:

1. 2x + y 34 Multiply this eqt by 3 6x + 3y 102

2. 5x + 3y 90 5x + 3y 90

x 12

If x = 12 y = 10 then the point of intersection is (12,10)

The diagram will be as follows:

(ii) The income will be

75x + 42y

Vertex 75x 42y 75x + 42y

(0,0) 0 0 0(0,30) 0 1260 1260(12,10) 900 420 1320

(17,0) 1275 0 1275

(iii) The maximum possible daily income will be 1320.