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A luxury bus has 100 units of seating area. There are two types of seats, Ordinary and First Class. Let the number of Ordinary seats be x and First Class seats be y.

Mathematics

ScanSolve AI Answer

Step-by-step solution

6 steps

Okay, Mamo💕, let's break down how to draw the graph for the inequalities from parts (a), (b), and (c), and then identify the feasible region.

First, let's list the inequalities you need to graph:

  • From (a): 2x+3y≤2002x + 3y \le 200
  • From (b): y≥10y \ge 10
  • From (c): x≥2yx \ge 2y
  • Also, since xx and yy represent the number of seats, we must have x≥0x \ge 0 and y≥0y \ge 0.
  1. Set up the Coordinate Axes

    • Draw a horizontal x-axis and a vertical y-axis on your graph paper.
    • The problem states to use a scale of 2 cm2 \text{ cm} to 10 units10 \text{ units} on both axes. This means every 2 cm2 \text{ cm} mark should be labeled as 10,20,30,…10, 20, 30, \dots.
    • For the x-axis, you'll need to go up to at least 100 (since 2x=200  ⟹  x=1002x = 200 \implies x=100). So, label from 0 to 100 or slightly beyond.
    • For the y-axis, you'll need to go up to at least 70 (since 3y=200  ⟹  y≈66.673y = 200 \implies y \approx 66.67). So, label from 0 to 70 or slightly beyond.
    • Label the x-axis as "Number of Ordinary Seats (xx)" and the y-axis as "Number of First Class Seats (yy)".
  2. Draw the Line for 2x+3y≤2002x + 3y \le 200

    • To draw the line, consider the equation 2x+3y=2002x + 3y = 200.
    • Find two points on this line:
      • If x=0x = 0, then 3y=200  ⟹  y=2003≈66.73y = 200 \implies y = \frac{200}{3} \approx 66.7. Plot the point (0,66.7)(0, 66.7).
      • If y=0y = 0, then 2x=200  ⟹  x=1002x = 200 \implies x = 100. Plot the point (100,0)(100, 0).
    • Draw a solid straight line connecting these two points.
    • Shading for 2x+3y≤2002x + 3y \le 200 (Unwanted Region): Test a point, e.g., (0,0)(0,0). 2(0)+3(0)=0≤2002(0) + 3(0) = 0 \le 200. This is true, so (0,0)(0,0) is in the wanted region. Therefore, shade the region above the line 2x+3y=2002x + 3y = 200.
  3. Draw the Line for y≥10y \ge 10}

    • To draw the line, consider the equation y=10y = 10.
    • This is a horizontal line passing through y=10y=10 on the y-axis.
    • Draw a solid horizontal line at y=10y=10.
    • Shading for y≥10y \ge 10 (Unwanted Region): Test a point, e.g., (0,0)(0,0). 0≥100 \ge 10. This is false, so (0,0)(0,0) is in the unwanted region. Therefore, shade the region below the line y=10y = 10.
  4. Draw the Line for x≥2yx \ge 2y}

    • To draw the line, consider the equation x=2yx = 2y (or y=12xy = \frac{1}{2}x).
    • Find two points on this line:
      • If x=0x = 0, then 0=2y  ⟹  y=00 = 2y \implies y = 0. Plot the point (0,0)(0,0).
      • If x=20x = 20, then 20=2y  ⟹  y=1020 = 2y \implies y = 10. Plot the point (20,10)(20,10).
      • If x=100x = 100, then 100=2y  ⟹  y=50100 = 2y \implies y = 50. Plot the point (100,50)(100,50).
    • Draw a solid straight line connecting these points.
    • Shading for x≥2yx \ge 2y (Unwanted Region): Test a point, e.g., (0,10)(0,10) (which is not on the line). 0≥2(10)  ⟹  0≥200 \ge 2(10) \implies 0 \ge 20. This is false, so (0,10)(0,10) is in the unwanted region. Therefore, shade the region above the line x=2yx = 2y (or to the left of it, if you consider the line from (0,0)(0,0) to (100,50)(100,50)).
  5. Shade for x≥0x \ge 0 and y≥0y \ge 0 (Unwanted Regions)

    • For x≥0x \ge 0: Shade the region to the left of the y-axis (where x<0x < 0).
    • For y≥0y \ge 0: Shade the region below the x-axis (where y<0y < 0).
  6. Identify the Feasible Region

    • The unshaded region on your graph is the feasible region, where all the inequalities are satisfied. This region represents all possible combinations of ordinary and first-class seats that meet the given conditions.
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