Applications of differentiation

Differentiation can be used to find the maximum or minimum value of a quantity — a process called optimisation. At a maximum or minimum point, the gradient (first derivative) equals zero, since the curve is momentarily flat. To confirm whether that point is a maximum or minimum, the second derivative test can be used: a negative second derivative indicates a maximum, and a positive second derivative indicates a minimum.

Key terms

Optimisation:

Using calculus to find a maximum or minimum value of a quantity.

Critical point:

A point where the derivative equals zero or is undefined.

Second derivative test:

A test using the second derivative to determine if a critical point is a maximum or minimum.

Questions

1. 1. Optimisation uses calculus to find:

   • A maximum or minimum value of a quantity
   • Only the total area of a shape
   • Nothing related to maximum or minimum values
   • The colour of a graph only

2. 2. At a maximum or minimum point, the gradient (first derivative):

   • Equals zero
   • Is always a large positive number
   • Is always a large negative number
   • Is undefined in every case

3. 3. A critical point is where:

   • The derivative equals zero or is undefined
   • The function is always increasing rapidly
   • The graph does not exist
   • The function has no derivative at all

4. 4. The second derivative test helps determine:

   • Whether a critical point is a maximum or minimum
   • The exact colour of a graph
   • Nothing related to maximum or minimum points
   • Only the starting value of a function

5. 5. A negative second derivative at a critical point indicates:

   • A maximum
   • A minimum
   • Neither a maximum nor minimum
   • An undefined function

6. 6. A positive second derivative at a critical point indicates:

   • A minimum
   • A maximum
   • Neither a maximum nor minimum
   • An undefined function

7. 7. At a maximum or minimum point, the curve is momentarily:

   • Flat
   • Extremely steep
   • Undefined
   • Moving at infinite speed

8. 8. If f(x) = -x² + 4x, then f'(x) = 0 when x equals:

   • 2
   • 4
   • 0
   • -2

9. 9. If f(x) = x² - 6x + 5, the critical point occurs at x equals:

   • 3
   • 6
   • 5
   • 0

10. 10. For f(x) = -x² + 4x, the second derivative f''(x) equals:

    • -2
    • 2
    • 4
    • 0

11. 11. Since f''(x) = -2 for f(x) = -x² + 4x, the critical point at x = 2 is a:

    • Maximum
    • Minimum
    • Neither maximum nor minimum
    • Point where the function is undefined

12. 12. Why is finding where the derivative equals zero the first step in solving an optimisation problem?

    • It identifies the critical points where a maximum or minimum could occur
    • This step has no relevance to solving optimisation problems
    • Derivatives are never used in optimisation, only original functions
    • Setting the derivative to zero always gives the final numerical answer immediately

13. 13. Why might a real-world optimisation problem (like maximising enclosed area) need to be expressed as a function of a single variable before differentiating?

    • Differentiation techniques for a single variable require the relationship to be reduced to one input before finding critical points
    • Functions with multiple variables are always simpler to differentiate directly
    • Reducing to one variable has no relevance to the optimisation process
    • Optimisation problems can never involve more than one variable from the start

14. 14. For a rectangular area problem, if the perimeter constraint gives width w = 50 − l, and area A = l(50 − l), the value of l that maximises area is:

    • 25
    • 50
    • 0
    • 100

15. 15. Why might optimisation problems in calculus often involve a constraint (like a fixed amount of fencing or material)?

    • A constraint provides the relationship needed to reduce the problem to a single variable before optimising
    • Constraints are never involved in real-world optimisation problems
    • A problem can always be optimised without any relationship between variables
    • Fixed constraints make optimisation problems impossible to solve using calculus

16. 16. Why might businesses use calculus-based optimisation to determine the price that maximises profit?

    • Profit can often be modelled as a function of price, and calculus finds the price where the rate of change of profit is zero
    • Price and profit have no mathematical relationship that calculus can model
    • Optimisation techniques are never applied to real business or economic problems
    • Maximising profit always requires guessing rather than any systematic calculation

17. 17. Why is checking the second derivative (or another method) necessary, rather than assuming every critical point is automatically a maximum?

    • A critical point could be a maximum, a minimum, or neither, so further testing is required to confirm which
    • Every critical point found by setting the derivative to zero is always automatically a maximum
    • The second derivative test provides no additional useful information about a critical point
    • Critical points never need any further testing once identified

18. 18. For a box with square base of side x and fixed volume, if the surface area function reduces to a single variable x, differentiating and solving for the critical point helps find:

    • The dimensions that minimise material used to build the box
    • The exact colour of the box
    • A value entirely unrelated to the box's dimensions
    • The volume of the box, ignoring surface area entirely

19. 19. If f(x) = x² - 8x + 12, the x-coordinate of the critical point is:

    • 4
    • 8
    • 12
    • 2

20. 20. Why might a company use optimisation to determine the order quantity that minimises total inventory cost (ordering cost plus storage cost)?

    • Balancing these competing costs mathematically identifies the quantity where total cost is lowest
    • Inventory costs can never be modelled or optimised using calculus
    • Ordering cost and storage cost always move in exactly the same direction with no trade-off
    • Optimisation techniques have no practical application to real business inventory decisions

21. 21. Why might a critical point at the very edge of a function's valid domain (rather than where the derivative equals zero) sometimes give the actual maximum or minimum in a real-world problem?

    • Real-world constraints can restrict the domain, meaning the best solution sometimes occurs at a boundary rather than a stationary point
    • Boundary points are never relevant to solving real-world optimisation problems
    • The derivative always equals zero at the true maximum or minimum in every scenario
    • Domain restrictions never affect where a genuine maximum or minimum can occur