Applications of Derivatives
Q. A cylindrical can is to be made with a fixed volume of 1000 cm³. What dimensions minimize the surface area? (2022) 2022
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A.
10, 10
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B.
5, 20
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C.
8, 15
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D.
6, 18
Solution
Using the formula for surface area and volume, the optimal dimensions are found to be radius = 5 and height = 20.
Correct Answer: C — 8, 15
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Q. A cylindrical can is to be made with a fixed volume of 1000 cm³. What dimensions minimize the surface area? (2022)
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A.
10 cm height, 10 cm radius
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B.
5 cm height, 15.87 cm radius
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C.
8 cm height, 12.5 cm radius
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D.
12 cm height, 8.33 cm radius
Solution
Using the formula for surface area and volume, the optimal dimensions are found to be 5 cm height and approximately 15.87 cm radius.
Correct Answer: B — 5 cm height, 15.87 cm radius
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Q. A cylindrical can is to be made with a volume of 1000 cm³. What dimensions minimize the surface area? (2021)
-
A.
10, 10
-
B.
5, 20
-
C.
8, 15
-
D.
6, 18
Solution
Using the formula for volume and surface area, the optimal dimensions are found to be radius = 5 cm and height = 20 cm.
Correct Answer: B — 5, 20
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Q. A farmer wants to fence a rectangular area of 200 m^2. What dimensions will minimize the fencing required? (2021)
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A.
10, 20
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B.
14, 14.28
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C.
15, 13.33
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D.
20, 10
Solution
For a fixed area, the minimum perimeter occurs when the rectangle is a square. Thus, dimensions are approximately 14 m by 14.28 m.
Correct Answer: B — 14, 14.28
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Q. A rectangle has a perimeter of 40 cm. What dimensions will maximize the area? (2022)
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A.
10 cm by 10 cm
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B.
15 cm by 5 cm
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C.
20 cm by 0 cm
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D.
12 cm by 8 cm
Solution
For maximum area, the rectangle should be a square. Thus, each side = 40/4 = 10 cm.
Correct Answer: A — 10 cm by 10 cm
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Q. A rectangle has a perimeter of 40 units. What dimensions maximize the area? (2022) 2022
-
A.
10, 10
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B.
5, 15
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C.
8, 12
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D.
6, 14
Solution
For maximum area, the rectangle should be a square. Thus, each side = 40/4 = 10.
Correct Answer: A — 10, 10
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Q. A rectangle has a perimeter of 40 units. What dimensions maximize the area? (2022)
-
A.
10, 10
-
B.
8, 12
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C.
6, 14
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D.
5, 15
Solution
For a fixed perimeter, the area is maximized when the rectangle is a square. Thus, side = 10 units.
Correct Answer: A — 10, 10
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Q. Determine the critical points of f(x) = 3x^4 - 8x^3 + 6. (2021)
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A.
(0, 6)
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B.
(1, 1)
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C.
(2, 0)
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D.
(3, -1)
Solution
f'(x) = 12x^3 - 24x^2. Setting f'(x) = 0 gives x = 0, 2. Check f(1) = 1.
Correct Answer: B — (1, 1)
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Q. Determine the critical points of f(x) = e^x - 2x. (2021)
Solution
f'(x) = e^x - 2. Setting f'(x) = 0 gives e^x = 2, so x = ln(2).
Correct Answer: B — 1
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Q. Determine the intervals where f(x) = -x^2 + 4x is concave up. (2023)
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A.
(-∞, 0)
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B.
(0, 2)
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C.
(2, ∞)
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D.
(0, 4)
Solution
f''(x) = -2, which is always negative, indicating concave down everywhere.
Correct Answer: C — (2, ∞)
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Q. Determine the intervals where f(x) = x^4 - 4x^3 has increasing behavior. (2023)
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A.
(-∞, 0)
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B.
(0, 2)
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C.
(2, ∞)
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D.
(0, 4)
Solution
f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3). f'(x) > 0 for x in (0, 3).
Correct Answer: B — (0, 2)
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Q. Determine the intervals where f(x) = x^4 - 4x^3 has local minima. (2020)
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A.
(0, 2)
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B.
(1, 3)
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C.
(2, 4)
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D.
(0, 1)
Solution
f'(x) = 4x^3 - 12x^2. Setting f'(x) = 0 gives x = 0, 3. Testing intervals shows local minima at (0, 2).
Correct Answer: A — (0, 2)
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Q. Determine the local maxima of f(x) = -x^3 + 3x^2 + 1. (2021)
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A.
(0, 1)
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B.
(1, 3)
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C.
(2, 5)
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D.
(3, 4)
Solution
f'(x) = -3x^2 + 6x. Setting f'(x) = 0 gives x = 0 or x = 2. f(2) = 5 is a local maximum.
Correct Answer: B — (1, 3)
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Q. Determine the local minima of f(x) = x^4 - 4x^2. (2021)
Solution
f'(x) = 4x^3 - 8x. Setting f'(x) = 0 gives x = 0, ±2. f(0) = 0.
Correct Answer: B — 0
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Q. Determine the maximum area of a triangle with a base of 10 units and height as a function of x. (2020)
Solution
Area = 1/2 * base * height = 5h. Max area occurs when h is maximized, thus Area = 50 when h = 10.
Correct Answer: B — 50
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Q. Determine the maximum height of the function f(x) = -x^2 + 6x + 5. (2020) 2020
Solution
The vertex occurs at x = 3. f(3) = -3^2 + 6*3 + 5 = 8.
Correct Answer: A — 8
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Q. Determine the maximum height of the projectile modeled by h(t) = -16t^2 + 64t + 80. (2020)
Solution
The maximum height occurs at t = -b/(2a) = 64/(2*16) = 2. h(2) = -16(2^2) + 64(2) + 80 = 80.
Correct Answer: A — 80
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Q. Determine the maximum value of f(x) = -x^2 + 6x - 8. (2022)
Solution
The maximum occurs at x = 3. f(3) = -3^2 + 6(3) - 8 = 6.
Correct Answer: C — 6
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Q. Determine the minimum value of f(x) = x^2 - 4x + 5. (2021)
Solution
The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4(2) + 5 = 1.
Correct Answer: A — 1
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Q. Determine the point of inflection for f(x) = x^4 - 4x^3 + 6. (2023)
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A.
(1, 3)
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B.
(2, 2)
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C.
(0, 6)
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D.
(3, 0)
Solution
f''(x) = 12x^2 - 24x. Setting f''(x) = 0 gives x(12x - 24) = 0, so x = 0 or x = 2. Check f(1) = 3.
Correct Answer: A — (1, 3)
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Q. Find the dimensions of a rectangle with a fixed area of 50 m^2 that minimizes the perimeter. (2021)
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A.
5, 10
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B.
7, 7.14
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C.
8, 6.25
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D.
10, 5
Solution
For a fixed area, the minimum perimeter occurs when the rectangle is a square. Thus, dimensions are approximately 7 m by 7.14 m.
Correct Answer: B — 7, 7.14
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Q. Find the dimensions of a rectangle with a fixed area of 50 square units that minimizes the perimeter. (2022) 2022
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A.
5, 10
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B.
7, 7.14
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C.
10, 5
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D.
8, 6.25
Solution
For minimum perimeter, the rectangle should be a square. Thus, side = sqrt(50) ≈ 7.07.
Correct Answer: B — 7, 7.14
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Q. Find the maximum area of a triangle with a base of 10 m and height varying. (2020)
Solution
Area = 1/2 * base * height. Max area occurs when height is maximized, thus Area = 1/2 * 10 * 10 = 50.
Correct Answer: B — 50
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Q. Find the maximum area of a triangle with a base of 10 units and height as a function of the base. (2021)
Solution
Area = 1/2 * base * height. Max area occurs when height is maximized at 10 units, giving Area = 50.
Correct Answer: B — 50
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Q. Find the maximum height of the projectile modeled by h(t) = -16t^2 + 32t + 48. (2020)
Solution
The maximum occurs at t = -b/(2a) = -32/(2*-16) = 1. h(1) = 64.
Correct Answer: A — 48
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Q. Find the maximum height of the projectile modeled by h(t) = -16t^2 + 64t + 48. (2020)
Solution
The maximum occurs at t = -b/(2a) = 64/(2*16) = 2. h(2) = -16(2^2) + 64(2) + 48 = 80.
Correct Answer: B — 64
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Q. Find the maximum value of the function f(x) = -2x^2 + 8x - 3. (2021) 2021
Solution
The function is a downward-opening parabola. The maximum occurs at x = -b/(2a) = -8/(2*-2) = 2. f(2) = -2(2^2) + 8(2) - 3 = 8.
Correct Answer: B — 8
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Q. Find the minimum value of f(x) = x^2 - 4x + 7. (2021)
Solution
The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4*2 + 7 = 3.
Correct Answer: A — 3
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Q. Find the minimum value of f(x) = x^2 - 4x + 7. (2021) 2021
Solution
The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4*2 + 7 = 3.
Correct Answer: A — 3
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Q. Find the minimum value of the function f(x) = 2x^2 - 8x + 10. (2022)
Solution
The minimum occurs at x = 2. f(2) = 2(2^2) - 8(2) + 10 = 6.
Correct Answer: B — 4
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