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Algebra - Linear programming, including inequalities-extended

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Linear inequalities define regions on a coordinate plane. An inequality like ax+by≤cax + by \leq c represents all points on one side of the line ax+by=cax + by = c, including the line itself.

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For inequalities involving << or >>, the boundary line is drawn as a dashed line. For ≤\leq or ≥\geq, a solid line is used.

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The 'Feasible Region' is the set of all points (x,y)(x, y) that satisfy all given inequalities (constraints) simultaneously. In IB Extended math, this region is often shaded or identified to solve optimization problems.

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Linear Programming involves finding the maximum or minimum value of an objective function, typically in the form P=ax+byP = ax + by, within the feasible region.

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The Fundamental Theorem of Linear Programming states that the optimal solution (maximum or minimum) occurs at one of the vertices (corners) of the feasible region.

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Non-negativity constraints, such as x≥0x \geq 0 and y≥0y \geq 0, are common in real-world problems to represent quantities that cannot be negative.

📐Formulae

ax+by≤cax + by \leq c

y≥mx+cy \geq mx + c

P=ax+by (Objective Function)P = ax + by \text{ (Objective Function)}

m=y2−y1x2−x1 (Gradient of boundary lines)m = \frac{y_2 - y_1}{x_2 - x_1} \text{ (Gradient of boundary lines)}

💡Examples

Problem 1:

A small factory produces two products, xx and yy. The constraints are given by: x+y≤10x + y \leq 10, x≥2x \geq 2, and y≥3y \geq 3. Find the feasible region and maximize the profit function P=4x+5yP = 4x + 5y.

Solution:

First, identify the vertices of the feasible region by finding the intersection points of the boundary lines:

  1. Intersection of x=2x = 2 and y=3y = 3: Point A(2,3)A(2, 3)
  2. Intersection of x=2x = 2 and x+y=10x + y = 10: 2+y=10  ⟹  y=82 + y = 10 \implies y = 8. Point B(2,8)B(2, 8)
  3. Intersection of y=3y = 3 and x+y=10x + y = 10: x+3=10  ⟹  x=7x + 3 = 10 \implies x = 7. Point C(7,3)C(7, 3)

Now, evaluate P=4x+5yP = 4x + 5y at each vertex:

  • At A(2,3)A(2, 3): P=4(2)+5(3)=8+15=23P = 4(2) + 5(3) = 8 + 15 = 23
  • At B(2,8)B(2, 8): P=4(2)+5(8)=8+40=48P = 4(2) + 5(8) = 8 + 40 = 48
  • At C(7,3)C(7, 3): P=4(7)+5(3)=28+15=43P = 4(7) + 5(3) = 28 + 15 = 43

The maximum profit is 48 at point (2,8)(2, 8).

Explanation:

We first convert the inequalities into boundary equations to find the corners of the shape formed by the constraints. Since the maximum must occur at a corner, we plug those coordinates into the profit formula and compare the results.

Problem 2:

Show the region satisfied by the inequality 2x−3y>62x - 3y > 6.

Solution:

  1. Find the boundary line by treating it as an equation: 2x−3y=62x - 3y = 6.
  2. Find intercepts:
    • If x=0x = 0, −3y=6  ⟹  y=−2-3y = 6 \implies y = -2. Point: (0,−2)(0, -2)
    • If y=0y = 0, 2x=6  ⟹  x=32x = 6 \implies x = 3. Point: (3,0)(3, 0)
  3. Draw a dashed line through (0,−2)(0, -2) and (3,0)(3, 0) because the inequality is strict (>>).
  4. Test a point not on the line, e.g., (0,0)(0, 0): 2(0)−3(0)>6  ⟹  0>62(0) - 3(0) > 6 \implies 0 > 6 This is false, so shade the region that does not contain (0,0)(0, 0).

Explanation:

The boundary line splits the plane into two halves. A test point helps determine which side satisfies the inequality. A dashed line is essential for '>>' because points on the line are not included in the solution set.