Application of Derivatives Solved Examples (12 Mathematics)
Applications of derivatives include finding maxima, minima, rate of change, and optimization problems. These examples cover critical points, second derivat
TL;DR: Applications of derivatives include finding maxima, minima, rate of change, and optimization problems. These examples cover critical points, second de…
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Applications of derivatives include finding maxima, minima, rate of change, and optimization problems. These examples cover critical points, second derivat
Application of Derivatives — Solved Numerical Examples (Step by Step)
Example 1: Find the critical points of f(x) = x^3 - 3x^2 + 2.
Solution: Step 1: Find first derivative.
f'(x) = 3x^2 - 6x
Step 2: Set f'(x) = 0 to find critical points.
3x^2 - 6x = 0
3x(x - 2) = 0
x = 0 or x = 2
Critical points are x = 0 and x = 2.
Example 2: For f(x) = x^3 - 3x^2 + 2, determine if x = 0 and x = 2 are maxima or minima using second derivative test.
Solution: Step 1: Find second derivative.
f'(x) = 3x^2 - 6x
f''(x) = 6x - 6
Step 2: Apply second derivative test.
At x = 0: f''(0) = 6(0) - 6 = -6 < 0
Since f''(0) < 0, x = 0 is a local maximum.
f(0) = 0 - 0 + 2 = 2
At x = 2: f''(2) = 6(2) - 6 = 12 - 6 = 6 > 0
Since f''(2) > 0, x = 2 is a local minimum.
f(2) = 8 - 12 + 2 = -2
Example 3: A rectangular box with open top is to be made from a piece of cardboard of dimensions 20 cm × 20 cm by cutting equal squares from each corner and folding. Find the dimensions that maximize volume.
Solution: Let x = side of square cut from each corner.
Height of box = x
Length of box = 20 - 2x
Width of box = 20 - 2x
Volume V = x(20 - 2x)^2
V = x(400 - 80x + 4x^2)
V = 400x - 80x^2 + 4x^3
Step 1: Find dV/dx.
dV/dx = 400 - 160x + 12x^2
Step 2: Set dV/dx = 0.
12x^2 - 160x + 400 = 0
3x^2 - 40x + 100 = 0
Using quadratic formula:
x = (40 ± √(1600 - 1200)) / 6 = (40 ± √400) / 6 = (40 ± 20) / 6
x = 60/6 = 10 or x = 20/6 = 10/3
Since 0 < x < 10, both values seem valid, but x = 10 gives zero length/width.
x = 10/3 cm
Dimensions: height = 10/3 cm, length = width = 20 - 20/3 = 40/3 cm
Maximum volume = (10/3) × (40/3) × (40/3) = 64,000/27 ≈ 2370.4 cm³
Example 4: The rate at which water flows out of a tank is proportional to the square root of the height of water. If dh/dt = -k√h where k is constant, find h as a function of t.
Solution: dh/dt = -k√h
Separate variables:
dh/√h = -k dt
Integrate both sides:
integral of h^(-1/2) dh = integral of -k dt
2√h = -kt + C
Using initial condition h = h0 at t = 0:
2√h0 = C
Solution: 2√h = -kt + 2√h0
√h = √h0 - kt/2
h = (√h0 - kt/2)²
Example 5: Find the equation of tangent line to y = x^2 + 3x at the point where x = 2.
Solution: Step 1: Find y-coordinate at x = 2.
y = (2)^2 + 3(2) = 4 + 6 = 10
Point: (2, 10)
Step 2: Find slope (derivative).
dy/dx = 2x + 3
At x = 2: dy/dx = 2(2) + 3 = 7
Step 3: Use point-slope form.
y - 10 = 7(x - 2)
y - 10 = 7x - 14
y = 7x - 4
Example 6: A company manufactures widgets at a cost of C(x) = 100 + 5x + 0.1x^2 where x is quantity. Find the production level that minimizes average cost.
Solution: Average cost AC = C(x) / x = (100 + 5x + 0.1x^2) / x
AC = 100/x + 5 + 0.1x
Step 1: Find dAC/dx.
dAC/dx = -100/x^2 + 0.1
Step 2: Set dAC/dx = 0.
-100/x^2 + 0.1 = 0
0.1 = 100/x^2
x^2 = 1000
x = √1000 ≈ 31.62 units
Step 3: Verify it's a minimum.
d²AC/dx² = 200/x^3 > 0 for x > 0, so it's a minimum.
Minimum AC = 100/31.62 + 5 + 0.1(31.62)
≈ 3.16 + 5 + 3.16 ≈ Rs 11.32 per widget
Example 7: Find the point on the curve y = √x closest to the point (4, 0).
Solution: Let point on curve be (x, √x).
Distance D = √[(x-4)² + (√x-0)²] = √[(x-4)² + x]
To minimize, minimize D² (avoids square root):
D² = (x-4)² + x = x² - 8x + 16 + x = x² - 7x + 16
Step 1: Find d(D²)/dx.
d(D²)/dx = 2x - 7
Step 2: Set equal to 0.
2x - 7 = 0
x = 7/2 = 3.5
y = √3.5 ≈ 1.87
Point on curve: (3.5, 1.87)
Distance = √[(3.5-4)² + 1.87²] = √[0.25 + 3.5] = √3.75 ≈ 1.94
Tips
- First derivative f'(x) = 0 gives critical points; check if max/min using second derivative test.
- f''(x) < 0 at critical point = local maximum; f''(x) > 0 = local minimum.
- For optimization problems, set up function, find critical points, check boundaries if finite domain.
- Tangent line slope = derivative at that point; use point-slope form y - y1 = m(x - x1).
Frequently Asked Questions
Why do we use the second derivative test?
The second derivative test determines whether a critical point is a maximum, minimum, or neither (inflection point). It avoids checking values on either side of the critical point.
What is the difference between local and global maximum?
A local (relative) maximum is higher than nearby points; global (absolute) maximum is the highest point on the entire domain. A function can have multiple local maxima but only one global maximum.
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