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A road contractor has to transport 240 tonnes of hardcore. He will use two types of lorries,  type A and type B. He has 3 type A lorries and 2 type B lorries. The capacity of each type A lorry is 8 tonnes while that of type B is 15 tonnes. All type A lorries must each make the same number of trips. Similarly all type B lorries must each make the same number of trips. The number of trips made by each type B lorry should be less than twice those made by each type A lorry. Each type A lorry must not make more than 6 trips.

  1. Take x to be the number of trips made by each type A lorry and y to represent the number of trips made by each type B lorry. Form all the inequalities in .x and y, to represent the above information. 
  2. On the grid provided, draw all the inequalities and shade the unwanted region. Take 1 cm to represent 1 unit on each of the axes. 
  3. The cost of operating each type A lorry is Ksh 5 000 per trip while that of operating cach type B lorry is Ksh 12 500 per trip. Determine the number of trips each type of lorry should make in order to minimize the cost of transporting the hardcore. Hence calculate the minimum cost. 

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  1. Y< 2X … i
    X ≤ 6 … ii
    8X + 15Y ≥ 240 … iii
    But Y=2y and X=3x
    y < 3x … i
    x ≤ 2 … ii
    4x + 5y ≥ 40 … iii
    x > 0 & y > 0
  2. y ≤ 3x … i
     x   0   4 
     y  0   12 

    4x + 5y ≥ 40 … iii
     x   0   10 
     y  8   0 
    Graph
  3. C= 5 000x + 12 500y
    (9, 2), (6, 4), (3, 6)
    5 000 x 9 + 12 500 x2 = 70 000
    c= 5000x 6 + 12 500x4 = 80 000
         5000x 3 + 12 500x6 = 90 000
    ∴ no of type A lorries = 9 
                type B lorries= 2

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