WAEC 2012 · Paper 2 · Q4

  1. (a)

    Indicate by shading graphically the set of all points P(x,y)P(x, y) in the OxyOxy plane that satisfy simultaneously the inequalities 2x−y≥−42x - y \ge -4, x+y≤10x + y \le 10, y−x>0y - x > 0, y≥2y \ge 2 and x≥0x \ge 0.

    Model answer
    246810246810xy(5, 5)y = 2x + 4x + y = 10y = xy = 2R

    Draw each boundary line, then shade the side that satisfies every inequality. The region RR has corners (5,5)(5, 5), (2,8)(2, 8), (0,4)(0, 4), (0,2)(0, 2), (2,2)(2, 2). The line y=xy = x is dashed because y−x>0y - x > 0 is strict (points on it are not included). WAEC accepts either shading the wanted region or shading the unwanted side, as long as you label the region clearly.

    For (b): check 10x+5y10x + 5y at each corner; the largest value, 75, is at (5,5)(5, 5).

  2. (b)

    Using the graph, find the values of xx and yy for which 10x+5y10x + 5y is maximum.

    Separate values with commas, e.g. 3, −2

Try it on a graph

The feasible region; test the corners in 10x + 5y.

Worked solution (try it first)

(a)

  1. Draw 2x−y=−42x - y = -4, x+y=10x + y = 10, y=xy = x (dashed, since y−x>0y - x > 0), y=2y = 2 and x=0x = 0, and shade the region that satisfies all five.
  2. Its corners: (0,2)(0, 2).
  3. y=2y = 2 and y=xy = x give (2,2)(2, 2).
  4. y=xy = x and x+y=10x + y = 10 give (5,5)(5, 5).
  5. x+y=10x + y = 10 and 2x−y=−42x - y = -4 give 3x=63x = 6, so (2,8)(2, 8).
  6. 2x−y=−42x - y = -4 and x=0x = 0 give (0,4)(0, 4).

(b)

  1. 10x+5y10x + 5y at the corners: 1010, 3030, 7575, 6060 and 2020.
  2. The maximum is 7575, at x=5x = 5, y=5y = 5.

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