Applications of Differentiation (Edexcel A-Level Mathematics): Revision Notes

• Differentiation is the process of finding the derivative of a function, which represents the rate of change. It is used to understand how a function behaves and to solve problems involving optimization.

• Optimization involves finding the maximum or minimum values of a function, which could represent the highest profit, lowest cost, shortest distance, or most efficient use of resources.

Step 1: Define the Problem

• Clearly understand and define what needs to be optimized (e.g., maximizing area, minimizing cost, etc.).
• Identify the variables involved and how they are related.

Step 2: Set Up the Function to Be Optimized

• Formulate the objective function, $f(x),$ which represents the quantity you want to maximize or minimize.
• Express the function in terms of a single variable, if possible.

Step 3: Differentiate the Function

• Find the first derivative $f'(x)$ of the objective function. This derivative represents the rate of change of the function with respect to the variable $x$.

Step 4: Find the Critical Points

• Solve $f'(x) = 0$ to find the critical points. These are the points where the function could have a maximum or minimum.
• Check where the derivative is undefined, as these points may also be critical.

Step 5: Classify the Critical Points

• Use the second derivative test to determine whether each critical point is a local maximum, local minimum, or neither:
• If $f''(x) > 0$, the function has a local minimum at that point.
• If $f''(x) < 0$, the function has a local maximum.
• If $f''(x) = 0$, the test is inconclusive, and further analysis is needed.

Step 6: Analyse the Results

• Determine the maximum or minimum value of the function based on the critical points.
• Consider the domain of the function and check any endpoints if the domain is restricted.

Example 1: Maximizing Area with a Fixed Perimeter

Problem: You have 100 meters of fencing to enclose a rectangular area. What dimensions should the rectangle have to maximize the enclosed area?

• Step 1: Define the Problem:
• Let the length of the rectangle be l and the width be w.
• The perimeter constraint is $2l + 2w = 100, or l + w = 50.$

• Step 2: Set Up the Function to Be Optimized:
• The area $A$ of the rectangle is $A = l \times w.$
• Using the perimeter constraint, express $w = 50 - l.$
• Substitute into the area function: $A(l) = l(50 - l) = 50l - l^2.$

• Step 3: Differentiate the Function:
• Find the first derivative: $A'(l) = 50 - 2l$.

• Step 4: Find the Critical Points:
• Set the derivative equal to zero: $50 - 2l = 0, which\ gives\ l = 25.$

• Step 5: Classify the Critical Points:
• Find the second derivative: $A''(l) = -2.$
• Since $A''(l) < 0$, the function has a local maximum at $l = 25.$

• Step 6: Analyse the Results:
• The rectangle should have dimensions $l = 25$meters and $w = 50 - 25 = 25$ meters to maximize the area.
• The maximum area is A = 25 × 25 = 625 square meters.

Example 2: Minimizing Cost

Problem: A company needs to manufacture a cylindrical can that holds 1 litre (1000 cm³) of liquid. The cost of the material for the top and bottom is higher than for the side. Find the dimensions that minimize the cost of the material.

• Step 1: Define the Problem:
• Let the radius of the base be r and the height of the cylinder be h.
• The volume constraint is $\pi r^2 h = 1000.$
• The cost function $C(r)$ involves the surface area: $C = 2\pi r^2$ (for the top and bottom) plus $2\pi r h$ (for the side).

• Step 2: Set Up the Function to Be Optimized:
• Express h from the volume constraint:$h = \frac{1000}{\pi r^2}.$
• Substitute into the cost function: $C(r) = 2\pi r^2 + 2\pi r \cdot \frac{1000}{\pi r^2} = 2\pi r^2 + \frac{2000}{r}.$

• Step 3: Differentiate the Function:
• Find the first derivative: $C'(r) = 4\pi r - \frac{2000}{r^2}$.

• Step 4: Find the Critical Points:
• Set the derivative equal to zero: $4\pi r - \frac{2000}{r^2} = 0$ $4\pi r^3 = 2000$ $r^3 = \frac{2000}{4\pi} = \frac{500}{\pi} \quad \Rightarrow \quad r = \left(\frac{500}{\pi}\right)^{\frac{1}{3}}$
• Step 5: Classify the Critical Points:
• Use the second derivative test: $C''(r) = 4\pi + \frac{4000}{r^3}.$
• Since $C''(r) > 0$, the cost function $C(r)$ has a local minimum at this critical point.

• Step 6: Analyse the Results:
• The optimal radius is $r = \left(\frac{500}{\pi}\right)^{\frac{1}{3}}.$
• The corresponding height is $h = \frac{1000}{\pi r^2}.$
• These dimensions minimize the cost of the material for the can.

• Modelling with differentiation involves setting up a mathematical function that describes the problem, differentiating it to find the rate of change, and then using this information to find optimal solutions.
• Optimization is a key application, where critical points found via differentiation help identify the best solution to a problem.
• This approach is widely applicable across many fields, making it an essential tool in both theoretical and applied mathematics.

Examples:

1. $T = 7s^2 - 3 \Rightarrow \dfrac{dT}{ds}$
2. $N = 5n^2 + 2n + 2 \Rightarrow \dfrac{dN}{dn}$ Often in practical situations, the letters are not $x$ and $y$.

Example Problem: The surface area, $A \, \text{cm}^2$, of an expanding sphere of radius $r \, \text{cm}$ is given by $A = 4\pi r^2$. Find the rate of change of the area with respect to the radius at the instant when the radius is 6 cm.

Another Example: A sector of a circle has an area of 100 cm².

a) Show that the perimeter ($P$) of this sector is given by the formula:

b) Find the minimum value for the perimeter.

(b) Solution:

Differentiate to find the minimum:

Since $r$ represents a length, we take $\boxed {r = 10}$.

Q4 (Edexcel 6664, Jan 2012, Q8) Figure $3$ shows a flowerbed. Its shape is a quarter of a circle of radius $x$ meters with two equal rectangles attached to it along its radii. Each rectangle has a length equal to $x$ meters and width equal to $y$ meters.

Given that the area of the flowerbed is 4 m²,

(a) Show that

(b) Hence show that the perimeter $P$ meters of the flowerbed is given by the equation

Note: No $y$'s, so use (a) to eliminate $y$'s.

(c) Use calculus to find the minimum value of $P$.

(d) Find the width of each rectangle when the perimeter is a minimum. Give your answer to the nearest centimetre.

(b)

(c)

(d)