Constraint Systems

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Constraint Systems · Grade 9 · Worksheet 2

1. A rectangular garden has a length that is 3 meters more than twice its width. The area of the garden must be at least 54 square meters, and the perimeter cannot exceed 30 meters. Write a system of inequalities that represents these constraints, using w for width and l for length. Answer: l = 2w + 3, l × w ≥ 54, 2l + 2w ≤ 30 Solution: - \( w \) = width of the garden (in meters) - \( l \) = length of the garden (in meters) "Length is 3 meters more than twice its width" means: l = 2w + 3 "Area must be at least 54 square meters" means: l \times w \geq 54 Perimeter of a rectangle is \( 2l + 2w \).
   Full step-by-step solution

   Let's go step by step. --- **Step 1: Define variables** We are told: - \( w \) = width of the garden (in meters) - \( l \) = length of the garden (in meters) --- **Step 2: Translate the length condition** "Length is 3 meters more than twice its width" means: \[ l = 2w + 3 \] --- **Step 3: Translate the area condition** "Area must be at least 54 square meters" means: \[ l \times w \geq 54 \] --- **Step 4: Translate the perimeter condition** Perimeter of a rectangle is \( 2l + 2w \). "Perimeter cannot exceed 30 meters" means: \[ 2l + 2w \leq 30 \] We can simplify this by dividing through by 2: \[ l + w \leq 15 \] But the problem's given correct answer uses \( 2l + 2w \leq 30 \), so we can keep it in that form. --- **Step 5: Write the system of inequalities** The system is: \[ l = 2w + 3 \] \[ l \times w \geq 54 \] \[ 2l + 2w \leq 30 \] --- **Final answer:** l = 2w + 3, l × w ≥ 54, 2l + 2w ≤ 30

2. Ava is organizing a school fundraiser and can spend at most $180 on prizes. She wants to buy at least 15 prizes total, with no more than 8 of the expensive prizes that cost $12 each and at least 10 of the budget prizes that cost $7 each. Write a system of inequalities that represents these constraints, where x represents the number of expensive prizes and y represents the number of budget prizes. Answer: 12x + 7y ≤ 180, x + y ≥ 15, x ≤ 8, y ≥ 10, x ≥ 0 Solution: Write the cost constraint: Each expensive prize costs $12 and each budget prize costs $7, with a maximum budget of $180.
   Full step-by-step solution

   Step 1: Write the cost constraint: Each expensive prize costs $12 and each budget prize costs $7, with a maximum budget of $180. This gives: 12x + 7y ≤ 180 Step 2: Write the total quantity constraint: At least 15 prizes total are needed. This gives: x + y ≥ 15 Step 3: Write the constraint for expensive prizes: No more than 8 expensive prizes. This gives: x ≤ 8 Step 4: Write the constraint for budget prizes: At least 10 budget prizes. This gives: y ≥ 10 Step 5: Include non-negativity constraints: Since we cannot have negative prizes, we also need: x ≥ 0 and y ≥ 0 Step 6: Combine all constraints into the system: 12x + 7y ≤ 180, x + y ≥ 15, x ≤ 8, y ≥ 10, x ≥ 0 The complete system of inequalities is: 12x + 7y ≤ 180, x + y ≥ 15, x ≤ 8, y ≥ 10, x ≥ 0

3. A tech company is analyzing the profitability of their new software product. The revenue function is R(x) = -0.5x² + 60x, where x is the price per license in dollars. The cost function is C(x) = 20x + 800. The company wants the profit to be at least $1,000, and they cannot charge more than $80 per license due to market constraints. Write a system of inequalities that represents these business constraints. Answer: P(x) = -0.5x² + 40x - 800 ≥ 1000, x ≤ 80, x > 0 Solution: Define the profit function P(x) = R(x) - C(x) Substitute the given functions: P(x) = (-0.5x² + 60x) - (20x + 800) Simplify: P(x) = -0.5x² + 60x - 20x - 800 = -0.5x² + 40x - 800 The profit must be at least $1,000: -0.5x² + 40x - 800 ≥ 1000 The price cannot exceed $80: x ≤ 80 Price must be…
   Full step-by-step solution

   Step 1: Define the profit function P(x) = R(x) - C(x) Step 2: Substitute the given functions: P(x) = (-0.5x² + 60x) - (20x + 800) Step 3: Simplify: P(x) = -0.5x² + 60x - 20x - 800 = -0.5x² + 40x - 800 Step 4: The profit must be at least $1,000: -0.5x² + 40x - 800 ≥ 1000 Step 5: The price cannot exceed $80: x ≤ 80 Step 6: Price must be positive: x > 0 Step 7: The complete system is: -0.5x² + 40x - 800 ≥ 1000, x ≤ 80, x > 0

4. Kaia is designing a rectangular mural on a coordinate plane. The mural's bottom-left corner is at the origin (0,0) and its top-right corner is at (24, 18). She wants to place two rectangular photo frames inside the mural. The first frame must have its center (x, y) at least 3 units from each side of the mural. The second frame must have its center (p, q) at least 4 units from each side of the mural. Additionally, the distance between the two centers must be at least 10 units. Write a system of inequalities that represents all constraints on the possible locations of the two centers. Answer: x ≥ 3, x ≤ 21, y ≥ 3, y ≤ 15, p ≥ 4, p ≤ 20, q ≥ 4, q ≤ 14, sqrt((x-p)^2 + (y-q)^2) ≥ 10 Solution: The mural boundaries are x=0, x=24, y=0, y=18. For the first frame's center (x,y) to be at least 3 units from each side: x ≥ 3, x ≤ 24-3=21, y ≥ 3, y ≤ 18-3=15.
   Full step-by-step solution

   Step 1: The mural boundaries are x=0, x=24, y=0, y=18. For the first frame's center (x,y) to be at least 3 units from each side: x ≥ 3, x ≤ 24-3=21, y ≥ 3, y ≤ 18-3=15. Step 2: For the second frame's center (p,q) to be at least 4 units from each side: p ≥ 4, p ≤ 24-4=20, q ≥ 4, q ≤ 18-4=14. Step 3: The distance between the two centers must be at least 10 units, so using the distance formula: sqrt((x-p)^2 + (y-q)^2) ≥ 10. Step 4: The complete system is: x ≥ 3, x ≤ 21, y ≥ 3, y ≤ 15, p ≥ 4, p ≤ 20, q ≥ 4, q ≤ 14, sqrt((x-p)^2 + (y-q)^2) ≥ 10.

5. Aroha is organizing a school fundraiser with two types of tickets: student tickets cost $12 and adult tickets cost $18. She needs to raise at least $900 and can sell no more than 65 tickets total. Write the system of inequalities that represents these constraints, where x is the number of student tickets and y is the number of adult tickets. Answer: 12x + 18y ≥ 900, x + y ≤ 65, x ≥ 0, y ≥ 0 Solution: Identify the revenue constraint - Aroha needs at least $900, so 12x + 18y ≥ 900 Identify the ticket quantity constraint - No more than 65 tickets total, so x + y ≤ 65 Add non-negativity constraints since you can't have negative tickets, so x ≥ 0 and y ≥ 0 The complete system is: 12x + 18y ≥ 900,…
   Full step-by-step solution

   Step 1: Identify the revenue constraint - Aroha needs at least $900, so 12x + 18y ≥ 900 Step 2: Identify the ticket quantity constraint - No more than 65 tickets total, so x + y ≤ 65 Step 3: Add non-negativity constraints since you can't have negative tickets, so x ≥ 0 and y ≥ 0 Step 4: The complete system is: 12x + 18y ≥ 900, x + y ≤ 65, x ≥ 0, y ≥ 0

6. Aisha is designing a rectangular stage for a school play. The stage must have a perimeter of 50 meters. She wants the length to be at least 5 meters more than the width, and the area must be at least 126 square meters to accommodate all the performers. Write a system of inequalities that represents these constraints, using L for length and W for width. Answer: 2L + 2W = 50, L ≥ W + 5, L × W ≥ 126 Solution: The perimeter formula connects length and width, while inequality signs represent minimum requirements. Area constraints create nonlinear relationships that must be considered alongside linear perimeter conditions.
   Full step-by-step solution

   When modeling real-world constraints with inequalities, we translate geometric relationships into mathematical statements. The perimeter formula connects length and width, while inequality signs represent minimum requirements. Area constraints create nonlinear relationships that must be considered alongside linear perimeter conditions.