Applications of Derivatives

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Q1. 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.

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Q2. What is 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 = 10, giving area = 50.

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Q3. If the revenue function is R(x) = 50x - 0.5x^2, find the number of units that maximizes revenue. (2023)

Solution:

Max revenue occurs at x = -b/(2a) = -50/(2*-0.5) = 50.

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Q4. What is the maximum height of the projectile modeled by h(t) = -16t^2 + 64t + 48? (2021)

Solution:

The maximum height occurs at t = -b/(2a) = -64/(2*-16) = 2. h(2) = -16(2^2) + 64(2) + 48 = 80.

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Q5. A cylindrical can is to be made with a volume of 1000 cm³. What dimensions minimize the surface area? (2021)

Solution:

Using the formula for volume and surface area, the optimal dimensions are found to be radius = 5 cm and height = 20 cm.

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Q6. 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.

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Q7. What is the maximum area of a triangle with a base of 10 units and height as a function of the base? (2021)

Solution:

Area = 0.5 * base * height. Max area occurs when height is 10, giving Area = 0.5 * 10 * 10 = 50.

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Q8. 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.

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Q9. 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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Q10. What is the value of x where f(x) = x^3 - 3x has a local maximum? (2022)

Solution:

f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = ±1. f(1) = -2.