Class 12 Maths: Application of Derivatives – Maxima & Minima

Why do curves matter in everyday life?

Ever wondered how a roller‑coaster designer knows where the ride will be steepest? Or how a farmer decides the best spot to plant for maximum yield? All that comes down to finding the highest or lowest points of a curve – that’s what maxima and minima are all about.

💡 In Simple Words: A maximum is the highest point on a curve, a minimum is the lowest. By looking at how the curve slopes (its derivative), we can spot these peaks and valleys without drawing the whole graph.

What are Maxima and Minima?

A maximum (plural maxima) is a point where the function’s value is larger than all nearby values. A minimum (plural minima) is the opposite – the smallest nearby value. Think of a mountain peak (max) and a valley (min) on a hiking trail.

Why do we care?

In physics, the highest point of a projectile’s path tells us the maximum height. In economics, the lowest cost tells a business where to cut expenses. In every case, the answer comes from calculus.

How to Find Maxima and Minima using Derivatives

Finding these special points is a step‑by‑step recipe. First, we need the derivative – that’s the fancy word for the slope of the curve at each point. When the slope is zero, the curve is flat for an instant – a possible peak or valley.

Worked Example 1

Find the maximum and minimum of f(x)=x^3-3x^2+2 on the interval [0,3].

• Step 1: f'(x)=3x^2-6x.
• Step 2: Set 3x^2-6x=0 → 3x(x-2)=0 → x=0 or x=2.
• Step 3: f''(x)=6x-6. Plug x=0: f''(0)=-6 (negative) ⇒ max at x=0. Plug x=2: f''(2)=6 (positive) ⇒ min at x=2.
• Step 4: Evaluate endpoints: f(0)=2, f(3)=3^3-3·3^2+2=27-27+2=2.

So the highest value on [0,3] is 2 (at x=0 and x=3) and the lowest is f(2)=2^3-3·2^2+2=8-12+2=-2.

Worked Example 2 (with a constraint)

Suppose a box has a square base and total surface area 150 cm². Find dimensions that give maximum volume.

• Let side of base = x cm, height = h cm.
• Surface area: x² + 4xh = 150 → h = (150‑x²)/(4x).
• Volume V = x²h = x²(150‑x²)/(4x) = (150x‑x³)/4.
• First derivative: V' = (150‑3x²)/4. Set to zero → 150‑3x²=0 → x²=50 → x≈7.07 cm.
• Second derivative: V'' = (‑6x)/4 = -1.5x, which is negative at x≈7.07, so we have a maximum.
• Height h = (150‑50)/(4·7.07) ≈ 2.24 cm.

Thus the box with a base side about 7.07 cm and height 2.24 cm holds the most volume under the given surface area.

Quick Summary

• Find f'(x) → locate where slope = 0.
• Solve f'(x)=0 → critical points.
• Use f''(x) to test:
  • f''>0 ⇒ minimum.
  • f''
• Don’t forget endpoints if the domain is closed.

Common Mistakes to Avoid

1. Ignoring points where the derivative does not exist – they can also be extrema.
2. Forgetting to test endpoints in a restricted interval – they often give the absolute max/min.
3. Assuming a zero second derivative means a flat point is automatically a max/min – you need a higher‑order test or sign change check.

📝 Likely Exam Questions

Q1. Find the local maximum and minimum of f(x)=2x^3‑9x^2+12x+1.

Answer: f'(x)=6x^2‑18x+12=6(x^2‑3x+2)=6(x‑1)(x‑2). Critical points x=1,2. f''(x)=12x‑18. f''(1)=-6 ⇒ max at x=1, f''(2)=6 ⇒ min at x=2.

Q2. A rectangular garden is to be fenced on three sides with 120 m of fencing. Find the dimensions that maximize the area.

Answer: Let length = x, width = y. Constraint: 2x + y = 120 → y = 120‑2x. Area A = xy = x(120‑2x) = 120x‑2x^2. A' = 120‑4x =0 → x=30 m, y=60 m. This gives maximum area.

Q3. Determine the absolute minimum of g(x)=x^4‑4x^3+6x^2 on [0,3].

Answer: g'(x)=4x^3‑12x^2+12x=4x(x^2‑3x+3). No real roots besides x=0. Check endpoints: g(0)=0, g(3)=81‑108+54=27. Minimum is 0 at x=0.

Q4. A company’s profit function is P(t)=‑t^3+12t^2‑36t+40, where t is months. When is profit highest?

Answer: P'(t)=‑3t^2+24t‑36=‑3(t^2‑8t+12)=‑3(t‑2)(t‑6). Critical points t=2,6. P''(t)=‑6t+24. P''(2)=12>0 ⇒ min, P''(6)=‑12