Application of derivatives — rate of change, tangent/normal, maxima/minima strategy

Question

Find the dimensions of a rectangle with perimeter 40 cm that has maximum area. Also, find the equation of the tangent to $y = x^3 - 3x + 2$ at $x = 1$ .

Solution — Step by Step

Let length = $x$ , breadth = $y$ . Perimeter: $2(x + y) = 40 \implies y = 20 - x$ .

Area $A = xy = x(20 - x) = 20x - x^2$

For maximum area: $\frac{dA}{dx} = 20 - 2x = 0 \implies x = 10$

Second derivative: $\frac{d^2A}{dx^2} = -2 < 0$ → confirms maximum.

So $x = 10$ , $y = 10$ . The rectangle is actually a square with side 10 cm. Maximum area = 100 sq cm.

$y = x^3 - 3x + 2$ . At $x = 1$ : $y = 1 - 3 + 2 = 0$ . Point is $(1, 0)$ .

Slope of tangent $= \frac{dy}{dx}\bigg|_{x=1} = (3x^2 - 3)\bigg|_{x=1} = 3 - 3 = 0$

Tangent equation: $y - 0 = 0(x - 1) \implies \mathbf{y = 0}$ (the x-axis).

Why This Works

graph TD
    A["Derivative Application: What is asked?"] --> B["Rate of change?"]
    A --> C["Tangent/Normal equation?"]
    A --> D["Maxima/Minima?"]
    B --> E["dy/dx at the given point"]
    C --> F["Find dy/dx = slope m at point"]
    C --> G["Tangent: y - y₁ = m x - x₁"]
    C --> H["Normal: y - y₁ = -1/m times x - x₁"]
    D --> I["Step 1: Find dy/dx = 0 → critical points"]
    D --> J["Step 2: Check d²y/dx²"]
    J --> K["d²y/dx² < 0 → Maximum"]
    J --> L["d²y/dx² > 0 → Minimum"]
    J --> M["d²y/dx² = 0 → Use first derivative test"]

The derivative gives us the instantaneous rate of change. For geometric interpretation, it is the slope of the tangent line. For optimization, critical points (where derivative = 0 or undefined) are candidates for maxima/minima.

The second derivative test tells us the nature of the critical point: if the curve bends downward ( $f'' < 0$ ), it is a local maximum; if it bends upward ( $f'' > 0$ ), it is a local minimum.

Optimization problems are CBSE board favourites. Common setups: maximum area of rectangle inscribed in a circle, maximum volume of cylinder inscribed in a sphere, minimum surface area for a given volume. The approach is always the same: express the quantity to optimize in terms of one variable, differentiate, set to zero, verify with second derivative.

Alternative Method

For the rectangle problem, you could use AM-GM inequality: for fixed perimeter $2(x+y) = 40$ , the product $xy$ is maximum when $x = y$ (AM-GM equality condition). This avoids calculus entirely and is faster for JEE MCQs.

In general: among all rectangles with a given perimeter, the square has the maximum area. Among all shapes with a given perimeter, the circle has the maximum area. These facts alone can solve many optimization MCQs.

Common Mistake

Not checking boundary values in optimization. When a problem has constraints (like $0 < x < 20$ for our rectangle), the maximum or minimum could occur at the boundary, not just at critical points. Compare the function value at all critical points AND at the endpoints. The absolute maximum is the largest of these values. Students often find the critical point and stop without checking boundaries — this can give wrong answers for constrained problems.

Rate of change: $\frac{dy}{dx}$ at a point

Tangent at $(x_1, y_1)$ : $y - y_1 = f'(x_1)(x - x_1)$

Normal at $(x_1, y_1)$ : $y - y_1 = -\frac{1}{f'(x_1)}(x - x_1)$

Critical points: $f'(x) = 0$ or $f'(x)$ undefined

Second derivative test: $f''(x) < 0$ → max, $f''(x) > 0$ → min

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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