Q. A cylindrical can is to be made with a fixed volume of 1000 cm³. What dimensions minimize the surface area? (2022)
A.
10 cm height, 10 cm radius
B.
5 cm height, 15.87 cm radius
C.
8 cm height, 12.5 cm radius
D.
12 cm height, 8.33 cm radius
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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 fixed volume of 1000 cm³. What dimensions minimize the surface area? (2022) 2022
A.
10, 10
B.
5, 20
C.
8, 15
D.
6, 18
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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 volume of 1000 cm³. What dimensions minimize the surface area? (2021)
A.
10, 10
B.
5, 20
C.
8, 15
D.
6, 18
Show solution
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)
A.
10, 20
B.
14, 14.28
C.
15, 13.33
D.
20, 10
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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 maximize the area? (2022)
A.
10 cm by 10 cm
B.
8 cm by 12 cm
C.
5 cm by 15 cm
D.
6 cm by 14 cm
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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 cm. What dimensions will maximize the area? (2022)
A.
10 cm by 10 cm
B.
15 cm by 5 cm
C.
20 cm by 0 cm
D.
12 cm by 8 cm
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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
B.
5, 15
C.
8, 12
D.
6, 14
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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
C.
6, 14
D.
5, 15
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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. At what point does the function f(x) = x^3 - 3x^2 + 4 have a local minimum? (2020)
A.
(1, 2)
B.
(2, 1)
C.
(0, 4)
D.
(3, 0)
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Solution
To find local minima, we find f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x = 0 and x = 2. Checking the second derivative, f''(2) = 6 > 0, so (2, 1) is a local minimum.
Correct Answer: A — (1, 2)
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Q. At which point does the function f(x) = -x^3 + 3x^2 + 4 have a local maximum? (2023)
A.
(0, 4)
B.
(1, 6)
C.
(2, 5)
D.
(3, 4)
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Solution
Setting f'(x) = -3x^2 + 6x = 0 gives x = 0 and x = 2. f''(2) < 0 indicates a local maximum at (2, 5).
Correct Answer: B — (1, 6)
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Q. Determine the critical points of f(x) = 3x^4 - 8x^3 + 6. (2021)
A.
(0, 6)
B.
(1, 1)
C.
(2, 0)
D.
(3, -1)
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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)
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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)
A.
(-∞, 0)
B.
(0, 2)
C.
(2, ∞)
D.
(0, 4)
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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^3 - 3x is increasing. (2021)
A.
(-∞, -1)
B.
(-1, 1)
C.
(1, ∞)
D.
(-∞, 1)
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Solution
f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = -1, 1. f'(x) > 0 for x > 1.
Correct Answer: C — (1, ∞)
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Q. Determine the intervals where f(x) = x^4 - 4x^3 has increasing behavior. (2023)
A.
(-∞, 0)
B.
(0, 2)
C.
(2, ∞)
D.
(0, 4)
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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)
A.
(0, 2)
B.
(1, 3)
C.
(2, 4)
D.
(0, 1)
Show solution
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)
A.
(0, 1)
B.
(1, 3)
C.
(2, 5)
D.
(3, 4)
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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 maxima of f(x) = x^4 - 8x^2 + 16. (2021)
A.
(0, 16)
B.
(2, 12)
C.
(4, 0)
D.
(1, 9)
Show solution
Solution
Find f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. f(2) = 12 is a local maximum.
Correct Answer: B — (2, 12)
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Q. Determine the local minima of f(x) = x^3 - 3x + 2. (2021)
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Solution
f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = 1. f(1) = 0.
Correct Answer: B — 0
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Q. Determine the local minima of f(x) = x^4 - 4x^2. (2021)
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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)
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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
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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 given by h(t) = -16t^2 + 64t + 80. (2023)
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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 height of the projectile modeled by h(t) = -16t^2 + 64t + 80. (2020)
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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)
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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)
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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)
A.
(1, 3)
B.
(2, 2)
C.
(0, 6)
D.
(3, 0)
Show solution
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. Determine the point where the function f(x) = 4x - x^2 has a maximum. (2022)
A.
(0, 0)
B.
(2, 4)
C.
(1, 3)
D.
(3, 3)
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Solution
The maximum occurs at x = 2, found by setting f'(x) = 4 - 2x = 0. f(2) = 4(2) - (2^2) = 4.
Correct Answer: B — (2, 4)
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Q. Find the critical points of f(x) = x^4 - 8x^2 + 16. (2021)
A.
(0, 16)
B.
(2, 0)
C.
(4, 0)
D.
(1, 15)
Show solution
Solution
f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. f(2) = 0 is a critical point.
Correct Answer: B — (2, 0)
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Q. Find the critical points of the function f(x) = x^4 - 8x^2 + 16. (2019)
A.
(0, 16)
B.
(2, 0)
C.
(4, 0)
D.
(1, 9)
Show solution
Solution
Setting f'(x) = 4x^3 - 16x = 0 gives x = 0, ±2. Evaluating f(2) = 0 shows (2, 0) is a critical point.
Correct Answer: B — (2, 0)
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