Applications of Derivatives

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Q1. A farmer wants to fence a rectangular field with 100 m of fencing. What dimensions will maximize the area? (2022)

Solution:

For maximum area, the rectangle should be a square. Each side = 100/4 = 25 m.

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Q2. A rectangle has a perimeter of 40 cm. What dimensions maximize the area? (2022)

Solution:

For maximum area, the rectangle should be a square. Thus, each side = 40/4 = 10 cm.

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Q3. Find the maximum area of a triangle with a base of 10 units and height as a function of x. (2022)

Solution:

Area = 1/2 * base * height. Max area occurs when height is maximized, which is 10 units, giving Area = 50.

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Q4. 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.

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Q5. Determine the intervals where f(x) = -x^2 + 4x is concave up. (2023)

Solution:

f''(x) = -2, which is always negative, indicating concave down everywhere.

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Q6. Determine the local maxima of f(x) = -x^3 + 3x^2 + 1. (2021)

Solution:

f'(x) = -3x^2 + 6x. Setting f'(x) = 0 gives x = 0 or x = 2. f(2) = 5 is a local maximum.

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Q7. A rectangle has a perimeter of 40 units. What dimensions maximize the area? (2022) 2022

Solution:

For maximum area, the rectangle should be a square. Thus, each side = 40/4 = 10.

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Q8. 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.

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Q9. Find the dimensions of a rectangle with a fixed area of 50 square units that minimizes the perimeter. (2022) 2022

Solution:

For minimum perimeter, the rectangle should be a square. Thus, side = sqrt(50) ≈ 7.07.

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Q10. What is the minimum value of f(x) = x^2 - 4x + 6? (2022)

Solution:

The vertex occurs at x = 2. f(2) = 2.