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A taxi company has “SUPER” taxis and “MINI” taxis. One morning a group of 45 people needs taxis.

For this group the taxi company uses x “SUPER” taxis and y “MINI” taxis. A “SUPER” taxi can carry 5 passengers and a “MINI” taxi can carry 3 passengers. So 5x + 3y 45. (a) The taxi company has 12 taxis. Write down another inequality in x and y to show this information. (b) The taxi company always uses at least 4 “MINI” taxis. Write down an inequality in y to show this information. (c) Draw x and y axes from 0 to 15 using 1 cm to represent 1 unit on each axis. (d) Draw three lines on your graph to show the inequality 5x + 3y 45 and the inequalities from parts (a) and (b). Shade the unwanted regions. [6] (e) The cost to the taxi company of using a “SUPER” taxi is $20 and the cost of using a “MINI” taxi is $10. The taxi company wants to find the cheapest way of providing “SUPER” and “MINI” taxis for this group of people. Find the two ways in which this can be done. (f) The taxi company decides to use 11 taxis for this group. (i) The taxi company charges $30 for the use of each “SUPER” taxi and $16 for the use of each “MINI” taxi. Find the two possible total charges. (ii) Find the largest possible profit the company can make, using 11 taxis. 22.03.2015

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Group of 45 persons. x super taxis and y mini taxis are used. capacity of a super taxi = 5 and of a mini taxi = 3

Total number of passengers: 5 x + 3 y ≥ 45 --- (1) a) Total number of taxis : N = 12 => x + y ≤ 12

b) y ≥ 4

c) see diagram attached. d) see diagram attached. The region above the line B is representation of 5 x + 3 y >= 45 The region below the line C (red) is representing x + y <= 12 The region to the right of vertical line D represents y >= 4

shading of unwanted regions. e) cost of using a super taxi : $20 and of a mini taxi = $10. Total cost of using x super taxis and y mini taxis = Cost = $ (20 x + 10 y)

To find the cheapest cost, we need to calculate the Cost on the boundaries of the unshaded- white region in the graph. The region of interest is the interior of triangle PQR. We evaluate this function Cost at P, Q and R:

At P: x = 6.6 y = 4 Cost = $ (20 * 6.6 + 10 * 4) = $ 172 At Q : x = 8 y = 4 Cost = $ 160 + 40 = $ 160 AT R : x = 4.5 y = 7.5 Cost = $ 90 + 75 = $ 165

Minimum cost is at Q , when 8 super taxis and 4 mini taxis are used.

2nd way: we draw a line on graph representing the cost. It is the green line. 20 x + 10 y = K = constant. We move this line parallel to itself. The point it touches the triangle PQR is the lowest cost or the highest cost. We find that.

f) x + y = 11 In that case, draw a line (black thin on the graph) x + y = 11. It meets the earlier lines y >= 4 and 5x + 3y >= 45 at point S : x = 7 and y = 4 and also at T (x = 6 and y = 5)

Cost at S = 20 x + 10 y = $ 140 + 40 = $ 180 Cost at T : = 20 x + 10 y = $ 120 + 50 = $ 170. In this case the minimum cost point is at P. But as x = 6.6, a fraction, we make x = 7. So we go to point S with x = 7 and y = 4. Cost is $ 180.

The minimum cost will be at point T with $ 170. with x = 6 and y = 5.

i) Revenue from passengers through charges = $ 30 x + 16 y

Revenue at point S: x = 7 , y = 4: = $ 210 + 64 = $ 274 Revenue at point T : x = 6 , y = 5 : = $ 180 + 80 = $ 260

Revenue at point S is more.

ii) Profit = Revenue - Cost : At point S: $ 274 - $180 = $ 94 at point T : $ 260 - $ 170 = $ 90

Hence at point S, the company makes more profit. That is with 11 taxis being used, choose 7 super taxis and 4 mini taxis to give maximum profit.