Mathematical Modelling — Telangana (SSC) Class 10 Mathematics Solutions (Free)
Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Mathematical Modelling" — important questions with detailed answers, download PDF for board exam preparation.
TL;DR: Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Mathematical Modelling" — important questions with detailed answers, download PD…
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Q1: A farmer has 100 meters of fencing to enclose a rectangular garden. If one side is x meters, express the area as a function of x and find dimensions for maximum area.
Step 1: Let length = x, width = w
Step 2: Perimeter = 2(x + w) = 100
Step 3: x + w = 50, so w = 50 - x
Step 4: Area A(x) = x × w = x(50 - x) = 50x - x²
Step 5: For maximum area, dA/dx = 50 - 2x = 0
Step 6: x = 25 meters
Step 7: w = 50 - 25 = 25 meters
Step 8: Maximum area = 25 × 25 = 625 m²
Final Answer: A(x) = 50x - x²; Maximum area when length = width = 25 m (square); Area = 625 m²
Q2: A ball is thrown upward from ground with initial velocity 40 m/s. Height h(t) = 40t - 5t² where t is time in seconds. Find maximum height and time to reach it.
Step 1: h(t) = 40t - 5t²
Step 2: For maximum height, dh/dt = 40 - 10t = 0
Step 3: t = 4 seconds
Step 4: Maximum height h(4) = 40(4) - 5(4)² = 160 - 80 = 80 meters
Step 5: Ball returns to ground when h(t) = 0
Step 6: 40t - 5t² = 0, t(40 - 5t) = 0
Step 7: t = 0 or t = 8 seconds
Final Answer: Maximum height = 80 m; Time to reach = 4 seconds; Returns to ground at t = 8 seconds
Q3: The cost of manufacturing x units is C(x) = 50 + 20x. Revenue from selling x units is R(x) = 40x. Find break-even point and profit function.
Step 1: Cost function C(x) = 50 + 20x
Step 2: Revenue function R(x) = 40x
Step 3: Profit P(x) = R(x) - C(x) = 40x - (50 + 20x) = 20x - 50
Step 4: Break-even when Profit = 0
Step 5: 20x - 50 = 0
Step 6: x = 2.5 units
Step 7: At break-even: Revenue = Cost = 40(2.5) = 100
Final Answer: Profit function P(x) = 20x - 50; Break-even at 2.5 units; Break-even cost/revenue = 100
Q4: Population of a city grows exponentially. If initial population is 100,000 and growth rate is 5% per year, find population after 10 years.
Step 1: Initial population P₀ = 100,000
Step 2: Growth rate r = 5% = 0.05 per year
Step 3: Formula P(t) = P₀(1 + r)ᵗ
Step 4: After t = 10 years: P(10) = 100,000 × (1.05)¹⁰
Step 5: (1.05)¹⁰ ≈ 1.6289
Step 6: P(10) = 100,000 × 1.6289 = 162,889
Final Answer: Population after 10 years ≈ 162,889
Q5: A rectangular box has a square base of side x cm. If volume is fixed at 500 cm³, express surface area as function of x.
Step 1: Square base with side x, let height = h
Step 2: Volume V = x²h = 500
Step 3: h = 500/x²
Step 4: Surface area SA = 2x² + 4xh (2 bases + 4 sides)
Step 5: SA(x) = 2x² + 4x(500/x²) = 2x² + 2000/x
Step 6: To minimize: dSA/dx = 4x - 2000/x² = 0
Step 7: 4x = 2000/x², 4x³ = 2000, x³ = 500
Step 8: x = ∛500 ≈ 7.94 cm
Final Answer: SA(x) = 2x² + 2000/x; Minimum surface area at x ≈ 7.94 cm
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