Optimisation through differentiation

optimisation is where we find the maximum or minimum value of a function, usually in order to solve a real-world problem - e.g. what is the optimal dimensions of a box to maximise volume, or what is the optimal price to charge to maximise profits.

Steps

1. Find an expression for the quantity to be optimised (e.g. area, volume, profit) in terms of just one variable.
   This often involves eliminating other variables using a limiting condition.
2. Find the derivative of the thing to be optimised with respect to the variable.
3. Find where the derivative is equal to zero - these are the critical points.
4. Use the second derivative to work out if each critical point is a maximum or minimum point.

Example: a wire has length 12 cm - what is the maximum area of a rectangle that can be formed?

• Let x = length
• Let y = width
• Area, A = xy
• Perimeter of rectangle = 2(x+y)=12
• Write y in terms of x:
  • 2(x+y) = 12
  • x + y = 6
  • y = 6 - x
• We can then substitute this expression for y into the area formula:
  • A = x(6 - x)
  • A = 6x - x^2
• We noww differentiate area with respect to x:
  • \frac{dA}{dx} = 6 - 2x
• To find the maximum area, we set the derivative equal to zero:
  • 6 - 2x = 0
  • 2x = 6
  • x = 3
• We can then find y:
  • y = 6 - x = 6 - 3 = 3 Find the area:
  • A = xy = 3 \times 3 = 9 Answer: the maximum area is found when both width=3 and length=3 (a square). The area is 9 cm^2.