Apply differentiation to practical problems involving maxima and minima in context

🏔️ What is a Maximum or Minimum?

Think of a mountain and a valley. The highest point on the mountain is a maximum – the function reaches its greatest value there. The lowest point in the valley is a minimum – the function reaches its smallest value. In maths, we call these points critical points because the slope (derivative) is zero or undefined.

🔍 How to Find Them?

1. Take the derivative: f'(x).
2. Set f'(x)=0 and solve for x (critical points).
3. Check the sign of f'(x) around each critical point:
   • if it changes from + to – → maximum
   • if it changes from – to + → minimum
4. Plug the x values back into f(x) to get the maximum/minimum values.

Step 1: Area function: A(x)=x(4x-x^2)=4x^2-x^3

Step 2: Derivative: A'(x)=8x-3x^2

Step 3: Set to zero: 8x-3x^2=0 \implies x(8-3x)=0 \implies x=0 \text{ or } x=\frac{8}{3}

Step 4: Test sign: for x<\frac{8}{3} → A'(x)>0, for x>\frac{8}{3} → A'(x)<0. So x=\frac{8}{3} is a maximum.

Step 5: Max area: A\!\left(\frac{8}{3}\right)=4\left(\frac{8}{3}\right)^2-\left(\frac{8}{3}\right)^3=\frac{128}{9}\approx14.22

Stopping time: set speed to zero: 20-4t=0 \implies t=5\text{ s}

Distance function: integrate speed: s(t)=\int v(t)\,dt=20t-2t^2+C

Assume car starts from rest at t=0, so C=0.

Distance travelled until stop: s(5)=20(5)-2(5)^2=100-50=50\text{ m}

💡 Examination Tips

• Always show your work – write the derivative, set it to zero, and test the sign.
• Check the domain of the function before solving; critical points outside the domain are irrelevant.
• Use the first derivative test for quick identification; the second derivative test is optional but can confirm the result.
• When dealing with real-world problems, translate the situation into a function first (e.g., area, distance, cost).
• Remember the units – they help verify that your answer makes sense.
• Practice with different types of functions (polynomials, trigonometric, exponential) to build confidence.