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Question of 94

Q.Solve the following system of inequalities graphically: x+2y≤10, x+y≥1, x−y≤0, x≥0, y≥0x + 2y \leq 10,\ x + y \geq 1,\ x - y \leq 0,\ x \geq 0,\ y \geq 0.

Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2022Subjective· 4mImportance★★★★★
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The five constraints together bound a quadrilateral region in the first quadrant with corners (0,1)(0,1), (0,5)(0,5), (103,103)\left(\dfrac{10}{3},\dfrac{10}{3}\right), and (0.5,0.5)(0.5,0.5).

The constraints are: x+2y≤10x+2y\leq10, x+y≥1x+y\geq1, x−y≤0x-y\leq0 (i.e. y≥xy\geq x), x≥0x\geq0, y≥0y\geq0.

Draw each boundary line and identify which side satisfies the inequality:

  • x+2y=10x+2y=10: shade the side containing the origin (since 0≤100\leq10), i.e. below/left of the line.
  • x+y=1x+y=1: shade the side away from the origin (since 0<10<1 fails), i.e. above/right of the line.
  • x=yx=y: shade the side where y≥xy\geq x (above this line).
  • x≥0, y≥0x\geq0,\ y\geq0: restrict to the first quadrant.

Find the vertices of the region common to all constraints by intersecting boundary lines pairwise (only where both original inequalities are satisfied):

x=0x=0 and x+y=1x+y=1: (0,1)(0,1)

x=0x=0 and x+2y=10x+2y=10: (0,5)(0,5)

x=yx=y and x+2y=10x+2y=10: substituting x=yx=y gives 3x=10⇒x=y=1033x=10\Rightarrow x=y=\dfrac{10}{3}, i.e. (103,103)\left(\dfrac{10}{3},\dfrac{10}{3}\right)

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