Chapter 3: Applications of the Derivative

In Chapter 2, we learned how to find the derivative ($f'(x)$) to determine the gradient of a curve. Now, we apply that skill to solve real-world problems. By finding where a gradient is zero, we can identify the highest and lowest points of a function—a process called Optimization.

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3.1 Stationary Points

A stationary point occurs where the gradient of the function is zero ($f'(x) = 0$). At these points, the tangent to the curve is perfectly horizontal.

There are three main types of stationary points:

1. Local Maximum: The “peak” of a hill. The gradient changes from positive to negative.
2. Local Minimum: The “bottom” of a valley. The gradient changes from negative to positive.
3. Stationary Point of Inflection: The curve flattens out momentarily but then continues in the same direction.

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3.2 Determining the Nature of a Point

To find out if a stationary point is a maximum or minimum, we use a Sign Table to check the gradient just before and just after the point.

$x$	$a−ϵ$ (Just before)	$a$ (The point)	$a+ϵ$ (Just after)
$f'(x)$	Positive (+)	$0$	Negative (-)
Shape	/	—	\
Conclusion		Local Maximum

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3.3 Increasing and Decreasing Functions

• A function is increasing on an interval if $f'(x) > 0$.
• A function is decreasing on an interval if $f'(x) < 0$.

Worked Example 1: Finding Stationary Points Find the coordinates and nature of the stationary points for $f(x) = x^3 – 3x$.

1. Differentiate: $f'(x) = 3x^2 – 3$.
2. Set to Zero: $3x^2 – 3 = 0 \rightarrow 3x^2 = 3 \rightarrow x^2 = 1 \rightarrow x = 1, -1$.
3. Find $y$-coordinates: > * $f(1) = (1)^3 – 3(1) = -2$. Point is $(1, -2)$.
   • $f(-1) = (-1)^3 – 3(-1) = 2$. Point is $(-1, 2)$.
4. Test Nature (for $x=1$):
   • Check $x=0$: $f'(0) = -3$ (Negative).
   • Check $x=2$: $f'(2) = 3(4)-3 = 9$ (Positive).
   • Since it goes Negative $\to$ Zero $\to$ Positive, $(1, -2)$ is a Local Minimum.

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3.4 Optimization in Practice

Optimization is the process of finding the “best” value (e.g., maximum area or minimum cost).

The Strategy:

1. Create an equation for the variable you want to optimize (e.g., Area $A$).
2. Ensure the equation has only one independent variable (you may need to substitute).
3. Differentiate the equation.
4. Set the derivative to zero and solve.

Worked Example 2: Maximizing Area

A farmer has 40m of fencing to create a rectangular paddock against a straight wall (only 3 sides need fencing). Find the dimensions that maximize the area.

1. Let the sides perpendicular to the wall be $x$. Let the side parallel be $y$.
2. Constraint: $2x + y = 40 \rightarrow y = 40 – 2x$.
3. Area Formula: $A = x \times y = x(40 – 2x) = 40x – 2x^2$.
4. Differentiate: $\frac{dA}{dx} = 40 – 4x$.
5. Set to Zero: $40 – 4x = 0 \rightarrow 4x = 40 \rightarrow x = 10$.
6. Find $y$: $y = 40 – 2(10) = 20$.
7. Answer: Dimensions are 10m by 20m for a maximum area of 200m².

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3.5 Practice Problems

Part A: Stationary Points

1. Find the stationary points of $f(x) = x^2 – 6x + 8$ and determine their nature.
2. For $y = 2x^3 – 9x^2 + 12x$, find the coordinates of the local maximum and local minimum.

Part B: Intervals

3. Determine the interval for which $f(x) = x^2 – 4x$ is a decreasing function.

Part C: Optimization

4. The height of a projectile is given by $h(t) = 30t – 5t^2$, where $t$ is time in seconds. Find the maximum height reached.
5. Challenge: A box with a square base and no top must have a volume of $32 \text{ cm}^3$. Find the side length of the base ($x$) that minimizes the surface area. (Hint: $V = x^2h$, $SA = x^2 + 4xh$).

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Solutions (Summary)

• 1. $f'(x) = 2x – 6 \rightarrow x = 3$. Point is $(3, -1)$. It is a Local Minimum.
• 2. $y’ = 6x^2 – 18x + 12 = 6(x-1)(x-2)$. Max at $(1, 5)$, Min at $(2, 4)$.
• 3. $f'(x) = 2x – 4$. Decreasing when $2x – 4 < 0 \rightarrow x < 2$.
• 4. $h'(t) = 30 – 10t = 0 \rightarrow t = 3$ seconds. $h(3) = 45 \text{m}$.
• 5. $h = \frac{32}{x^2}$. $SA = x^2 + 4x(\frac{32}{x^2}) = x^2 + \frac{128}{x}$. $SA’ = 2x – \frac{128}{x^2} = 0 \rightarrow x^3 = 64 \rightarrow x = 4 \text{ cm}$.

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