Q. Find the dimensions of a box with a square base that maximizes volume given a surface area of 600 sq. units. (2020)
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A.
10, 10
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B.
15, 15
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C.
12, 12
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D.
20, 20
Solution
Let x be the side of the base and h the height. The surface area constraint gives 2x^2 + 4xh = 600. Max volume occurs at x = 12.
Correct Answer: C — 12, 12
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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
-
A.
5, 10
-
B.
7, 7.14
-
C.
10, 5
-
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 dimensions of a rectangle with a fixed area of 50 square units that minimizes the perimeter. (2020)
-
A.
5, 10
-
B.
7, 7
-
C.
10, 5
-
D.
8, 6.25
Solution
For a fixed area, the perimeter is minimized when the rectangle is a square. Thus, side = √50.
Correct Answer: B — 7, 7
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Q. Find the local maxima of f(x) = -x^3 + 3x^2 + 1. (2020)
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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, 1)
Solution
f'(x) = -3x^2 + 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2. f(2) = 5.
Correct Answer: B — (1, 3)
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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 area of a triangle with a fixed perimeter of 30 cm. (2022)
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A.
75 cm²
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B.
100 cm²
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C.
50 cm²
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D.
60 cm²
Solution
For maximum area, the triangle should be equilateral. Area = (sqrt(3)/4) * (10)^2 = 75 cm².
Correct Answer: A — 75 cm²
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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 + 6. (2021)
Solution
The vertex form gives the minimum at x = 2. f(2) = 2.
Correct Answer: A — 2
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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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Q. Find the point of inflection for f(x) = x^3 - 6x^2 + 9x. (2022)
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A.
(1, 4)
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B.
(2, 3)
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C.
(3, 0)
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D.
(0, 0)
Solution
f''(x) = 6x - 12. Setting f''(x) = 0 gives x = 2. f(2) = 3.
Correct Answer: C — (3, 0)
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Q. Find the point on the curve y = x^3 - 3x^2 + 4 that has a horizontal tangent. (2023)
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A.
(0, 4)
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B.
(1, 2)
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C.
(2, 2)
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D.
(3, 4)
Solution
To find horizontal tangents, set the derivative y' = 3x^2 - 6x = 0. This gives x = 0 and x = 2. The point (1, 2) has a horizontal tangent.
Correct Answer: B — (1, 2)
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Q. Find the slope of the tangent line to f(x) = 2x^3 - 3x^2 + 4 at x = 1. (2021)
Solution
f'(x) = 6x^2 - 6. f'(1) = 6(1)^2 - 6 = 0.
Correct Answer: B — 2
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Q. Find the slope of the tangent line to f(x) = x^2 + 2x at x = 1. (2022)
Solution
f'(x) = 2x + 2. At x = 1, f'(1) = 4.
Correct Answer: A — 2
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Q. For the function f(x) = -x^2 + 4x + 1, find the x-coordinate of the vertex. (2023)
Solution
The vertex x-coordinate is given by -b/(2a) = -4/(2*-1) = 2.
Correct Answer: A — 2
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Q. For the function f(x) = 2x^3 - 9x^2 + 12x, find the critical points. (2022)
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A.
(0, 0)
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B.
(1, 5)
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C.
(2, 0)
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D.
(3, 3)
Solution
Set f'(x) = 0. f'(x) = 6x^2 - 18x + 12 = 0. Critical points are x = 2.
Correct Answer: C — (2, 0)
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Q. For the function f(x) = 2x^3 - 9x^2 + 12x, find the intervals of increase. (2022)
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A.
(-∞, 0)
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B.
(0, 3)
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C.
(3, ∞)
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D.
(0, 2)
Solution
f'(x) = 6x^2 - 18x + 12. The critical points are x = 1 and x = 2. Test intervals show increase in (0, 3).
Correct Answer: B — (0, 3)
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Q. For the function f(x) = 2x^3 - 9x^2 + 12x, find the local maxima. (2023) 2023
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A.
(1, 5)
-
B.
(2, 6)
-
C.
(3, 3)
-
D.
(0, 0)
Solution
Find f'(x) = 6x^2 - 18x + 12, set to 0. Local maxima at x = 2 gives f(2) = 6.
Correct Answer: B — (2, 6)
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Q. For the function f(x) = 3x^2 - 12x + 7, find the minimum value. (2022)
Solution
The vertex occurs at x = 2. f(2) = 3(2^2) - 12(2) + 7 = -5.
Correct Answer: B — -4
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Q. For the function f(x) = 3x^2 - 12x + 7, find the x-coordinate of the vertex. (2022)
Solution
The x-coordinate of the vertex is given by x = -b/(2a) = 12/(2*3) = 2.
Correct Answer: B — 2
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Q. For the function f(x) = 3x^2 - 12x + 9, find the vertex. (2021)
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A.
(2, 3)
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B.
(3, 0)
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C.
(0, 9)
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D.
(1, 6)
Solution
The vertex occurs at x = -b/(2a) = 12/(2*3) = 2. f(2) = 3(2^2) - 12(2) + 9 = 3.
Correct Answer: A — (2, 3)
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Q. For the function f(x) = 3x^2 - 12x + 9, find the x-coordinate of the vertex. (2021)
Solution
The vertex x-coordinate is found using -b/(2a) = 12/(2*3) = 2.
Correct Answer: B — 2
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Q. For the function f(x) = x^2 + 2x, find the local maximum. (2022)
Solution
f'(x) = 2x + 2. Setting f'(x) = 0 gives x = -1. f(-1) = 1.
Correct Answer: A — -1
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Q. If f(x) = x^3 - 3x^2 + 4, find the critical points. (2022)
-
A.
1, 2
-
B.
0, 3
-
C.
2, 4
-
D.
1, 3
Solution
f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2.
Correct Answer: A — 1, 2
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Q. If the cost function is C(x) = 3x^2 + 12x + 5, find the minimum cost. (2020)
Solution
The minimum cost occurs at x = -b/(2a) = -12/(2*3) = -2. C(-2) = 3(-2)^2 + 12(-2) + 5 = 8.
Correct Answer: B — 8
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