2-4 Applications on maximum and minimum values: practice questions

Correct answer: (C) 25 cm²

Explanation: Step 1: Perimeter = 20, so width x and length 10 − x; Area A = x(10 − x) = 10x − x² Step 2: Differentiate: A'(x) = 10 − 2x Step 3: Set A'(x) = 0: 10 − 2x = 0, so x = 5 Step 4: A''(x) = −2 less than 0, so x = 5 gives a maximum; it is a 5 cm × 5 cm square ∴ Largest area = 5 × 5 = 25 cm²

Correct answer: (D) 100

Explanation: Step 1: Let the numbers be x and 20 − x; product P = x(20 − x) = 20x − x² Step 2: Differentiate: P'(x) = 20 − 2x Step 3: Set P'(x) = 0: 20 − 2x = 0, so x = 10 Step 4: P''(x) = −2 less than 0, so x = 10 gives a maximum; the numbers are 10 and 10 ∴ Greatest product = 10 × 10 = 100

Correct answer: (C) x = 15

Explanation: Step 1: Let P(x) = x(30 − x) = 30x − x² Step 2: Differentiate: P'(x) = 30 − 2x Step 3: Set P'(x) = 0: 30 − 2x = 0, so x = 15 Step 4: P''(x) = −2 less than 0, so x = 15 gives a maximum ∴ x = 15

Correct answer: (A) 16

Explanation: Step 1: Let the base run from −x to x (width 2x) with top corners on the semicircle at height y = √(16 − x²) Step 2: Area A = 2x√(16 − x²); maximize A² = 4x²(16 − x²) = 64x² − 4x⁴ Step 3: Differentiate: (A²)' = 128x − 16x³; set = 0: 128x = 16x³, so x² = 8, x = √8 Step 4: Then y = √(16 − 8) = √8, giving a maximum; width 2√8, height √8 ∴ Maximum area = 2√8 × √8 = 2 × 8 = 16 square units

Correct answer: (D) 2 cm

Explanation: Step 1: Volume πr²h = 16π, so h = 16/r² Step 2: Total surface S = 2πr² + 2πrh = 2πr² + 2πr(16/r²) = 2πr² + 32π/r Step 3: Differentiate: S'(r) = 4πr − 32π/r²; set = 0: 4πr = 32π/r², so r³ = 8, r = 2 Step 4: S''(r) = 4π + 64π/r³ greater than 0, so r = 2 gives a minimum ∴ Base radius = 2 cm

Correct answer: (B) 108 cm³

Explanation: Step 1: Let base side = x and height = h; surface x² + 4xh = 108, so h = (108 − x²)/(4x) Step 2: Volume V = x²h = x² · (108 − x²)/(4x) = (108x − x³)/4 Step 3: Differentiate: V'(x) = (108 − 3x²)/4; set = 0: 108 − 3x² = 0, so x² = 36, x = 6 Step 4: V''(x) = −6x/4 less than 0 at x = 6, so it is a maximum; h = (108 − 36)/(24) = 3 ∴ Maximum volume = 6² × 3 = 108 cm³

Correct answer: (C) 32

Explanation: Step 1: By symmetry the corners are at (−x, 0), (x, 0) and top at height y = 12 − x²; width = 2x, height = 12 − x² Step 2: Area A = 2x(12 − x²) = 24x − 2x³ Step 3: Differentiate: A'(x) = 24 − 6x²; set = 0: x² = 4, x = 2 Step 4: A''(x) = −12x less than 0 at x = 2, so it is a maximum; height = 12 − 4 = 8, width = 4 ∴ Maximum area = 4 × 8 = 32 square units

Correct answer: (C) 20 cm

Explanation: Step 1: Let base side = x and height = h; volume x²h = 4000, so h = 4000/x² Step 2: Surface area S = x² + 4xh = x² + 4x(4000/x²) = x² + 16000/x Step 3: Differentiate: S'(x) = 2x − 16000/x²; set = 0: 2x = 16000/x², so x³ = 8000, x = 20 Step 4: S''(x) = 2 + 32000/x³ greater than 0, so x = 20 gives a minimum ∴ Base side length = 20 cm

Correct answer: (B) (7/2, √14/2)

Explanation: Step 1: D² = (x − 4)² + y² = (x − 4)² + x; d(D²)/dx = 2(x − 4) + 1 = 0 ⇒ x = 7/2 (second derivative 2 > 0, a minimum). Step 2: y = √(7/2) = √14/2. ∴ (7/2, √14/2)

Correct answer: (A) 8 and 16

Explanation: Step 1: Let the numbers be x and y with x + y = 24; maximize P = x·y², so x = 24 − y Step 2: P(y) = (24 − y)y² = 24y² − y³ Step 3: Differentiate: P'(y) = 48y − 3y² = 3y(16 − y); set = 0: y = 0 (rejected) or y = 16 Step 4: P''(y) = 48 − 6y; at y = 16, P'' = −48 less than 0, so y = 16 gives a maximum; then x = 24 − 16 = 8 ∴ The two numbers are 8 and 16