Apply differentiation to practical problems involving maxima and minima in context

Calculus – Applying Differentiation and Integration to Practical Problems (Cambridge IGCSE 0606)

1. Syllabus Overview

The calculus unit of Cambridge IGCSE Additional Mathematics (0606) is divided into two main parts:

• 14 – Differentiation (topics 14.1 – 14.15)
• 15 – Integration & Applications (topics 15.1 – 15.6)

The notes below follow the syllabus order, give concise theory, a systematic problem‑solving method, and worked examples that mirror exam style.

2. Differentiation

2.1 What is a Derivative?

The derivative of a function \(f(x)\) at a point \(x\) measures the instantaneous rate of change of \(f\) with respect to \(x\). Geometrically it is the gradient of the tangent line at that point.

Notation used in the syllabus:

• \(f'(x)\) – prime notation
• \(\displaystyle\frac{dy}{dx}\) – Leibniz notation (useful for rate‑of‑change problems)
• \(\displaystyle\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\) – limit definition (conceptual only)

2.2 Basic Differentiation Rules (14.1 – 14.3)

Function \(f(x)\)	Derivative \(f'(x)\)
\(c\) (constant)	0
\(x^{n}\) ( \(n\) rational)	\(n\,x^{\,n-1}\)
\(\sin x\)	\(\cos x\)
\(\cos x\)	\(-\sin x\)
\(\tan x\)	\(\sec^{2}x\)
\(e^{x}\)	\(e^{x}\)
\(a^{x}\) ( \(a>0\) )	\(a^{x}\ln a\)
\(\ln x\)	\(\dfrac1{x}\)

2.3 Product and Quotient Rules (14.4)

• Product rule: \((uv)' = u'v + uv'\)
• Quotient rule: \(\displaystyle\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^{2}}\)

2.4 Stationary Points, First‑ and Second‑Derivative Tests (14.6 – 14.9)

1. Find \(f'(x)\) and set \(f'(x)=0\) (or locate points where \(f'\) is undefined) – these are the stationary points.
2. First‑derivative sign test (useful when the second derivative is messy):
   • Choose a test value in each interval determined by the stationary points.
   • If \(f'\) changes from positive to negative, the stationary point is a local maximum.
   • If \(f'\) changes from negative to positive, it is a local minimum.

   Example: \(f(x)=x^{3}-3x\).
   \(f'(x)=3x^{2}-3=3(x^{2}-1)=3(x-1)(x+1)\).
   Stationary points at \(x=-1,\,1\).
   Sign of \(f'\): \((-∞,-1)\) (+), \((-1,1)\) (–), \((1,∞)\) (+).
   Hence \(x=-1\) is a local maximum, \(x=1\) a local minimum.

3. Second‑derivative test (preferred when \(f''\) is easy):
   • \(f''(x)>0\) ⇒ local minimum.
   • \(f''(x)<0\) ⇒ local maximum.
   • \(f''(x)=0\) ⇒ inconclusive – revert to the first‑derivative sign test.

2.5 Gradient, Tangent and Normal (14.5)

For a curve \(y=f(x)\) at a point \((x_{0},f(x_{0}))\):

• Gradient of the tangent: \(m_{\text{t}} = f'(x_{0})\)
• Equation of the tangent: \(y-f(x_{0}) = f'(x_{0})(x-x_{0})\)
• Gradient of the normal: \(m_{\text{n}} = -\dfrac{1}{f'(x_{0})}\) (provided \(f'(x_{0})eq0\))

2.6 Implicit Differentiation (optional, useful for curves such as circles)

If an equation involves both \(x\) and \(y\) and cannot be solved for \(y\) explicitly, differentiate each term with respect to \(x\) treating \(y\) as a function of \(x\) (apply \(\dfrac{dy}{dx}\) when differentiating a \(y\) term).

3. Integration

3.1 Anti‑derivatives (15.1)

If \(F'(x)=f(x)\) then \(F(x)\) is an anti‑derivative of \(f\). The general form includes a constant of integration \(C\):

\[ \int f(x)\,dx = F(x)+C \]

3.2 Integration Rules (reverse of differentiation) (15.2)

Integrand \(f(x)\)	Integral \(\displaystyle\int f(x)\,dx\)
\(x^{n}\) ( \(neq-1\) )	\(\dfrac{x^{n+1}}{n+1}+C\)
\(\dfrac1{x}\)	\(\ln|x|+C\)
\(\sin(ax+b)\)	\(-\dfrac{1}{a}\cos(ax+b)+C\)
\(\cos(ax+b)\)	\(\dfrac{1}{a}\sin(ax+b)+C\)
\(\sec^{2}(ax+b)\)	\(\dfrac{1}{a}\tan(ax+b)+C\)
\(e^{ax+b}\)	\(\dfrac{1}{a}e^{ax+b}+C\)
\((ax+b)^{n}\) ( \(neq-1\) )	\(\dfrac{(ax+b)^{n+1}}{a(n+1)}+C\)
\(a^{x}\)	\(\dfrac{a^{x}}{\ln a}+C\)

3.3 Definite Integrals and Area (15.3)

For a continuous function \(f\) on \([a,b]\):

\[ \int_{a}^{b} f(x)\,dx = F(b)-F(a) \]

where \(F\) is any anti‑derivative of \(f\). The value represents the signed area between the curve and the \(x\)-axis.

Area between two curves

If \(g(x)\ge h(x)\) on \([a,b]\) then

\[ \text{Area}= \int_{a}^{b}\bigl[g(x)-h(x)\bigr]\,dx . \]

Example: Find the area between \(y=x^{2}\) and \(y=x\) from \(x=0\) to \(x=1\).

\[ \int_{0}^{1}\bigl(x - x^{2}\bigr)\,dx = \Bigl[\tfrac12x^{2}-\tfrac13x^{3}\Bigr]_{0}^{1}= \tfrac12-\tfrac13 = \tfrac16\;\text{square units}. \]

3.4 Kinematics – Displacement, Velocity, Acceleration (15.4)

• If \(s(t)\) is displacement, then \(v(t)=\dfrac{ds}{dt}\) and \(a(t)=\dfrac{dv}{dt}\).
• Given any one of the three, integrate (adding the constant of integration using the given initial condition) to obtain the others.

Graph‑drawing requirement (all five graphs)

From a single expression (e.g. \(s(t)=5t^{2}-2t^{3}\)) the syllabus expects you to be able to sketch:

1. Displacement \(s\) vs. time \(t\)
2. Velocity \(v\) vs. time \(t\)
3. Acceleration \(a\) vs. time \(t\)
4. Velocity \(v\) vs. displacement \(s\)
5. Acceleration \(a\) vs. displacement \(s\)

Key steps: find stationary points, determine sign of \(v\) and \(a\), locate points where \(v=0\) (turning points of \(s\)), and use the relationships \(a=\dfrac{dv}{dt}=\dfrac{dv}{ds}\,v\) where required.

Worked kinematics example

Given \(a(t)=6-2t\) m s\(^{-2}\) with \(v(0)=4\) m s\(^{-1}\) and \(s(0)=0\):

1. Integrate acceleration: \(v(t)=\int(6-2t)\,dt = 6t - t^{2}+C\). Using \(v(0)=4\) gives \(C=4\). Hence \(v(t)=6t-t^{2}+4\).
2. Integrate velocity: \(s(t)=\int(6t-t^{2}+4)\,dt = 3t^{2}-\tfrac13t^{3}+4t +C\). With \(s(0)=0\) we get \(C=0\).
3. Stationary points of \(s\) occur when \(v=0\): solve \(6t-t^{2}+4=0\) ⇒ \(t=2\) s (the other root is negative). At \(t=2\), \(s=3(4)-\tfrac13(8)+8=12-2.\overline{6}+8=17.\overline{3}\) m.
4. Sketch the three \(t\)-graphs using the expressions above, then obtain the \(v\)–\(s\) and \(a\)–\(s\) curves by eliminating \(t\) (or by plotting points).

3.5 Work and Energy (optional, often appears in exam questions)

Work done by a variable force \(F(x)\) over \([a,b]\) is \(\displaystyle W=\int_{a}^{b}F(x)\,dx\).

4. Optimisation – Maxima and Minima in Context (14.6 – 14.9, 15.5)

4.1 Why Optimisation?

Real‑world problems frequently ask for the greatest or least possible value of a quantity: profit, cost, area, material, time, etc.

4.2 Systematic Procedure (7‑step method)

Step	What to do	Why
1	Translate the word problem into a single‑variable function \(f(x)\) representing the quantity to be optimised.	Creates a mathematical model.
2	State the domain (and any integer or positivity restrictions).	Ensures only feasible values are considered.
3	Differentiate: find \(f'(x)\).	Stationary points occur where \(f'(x)=0\) or undefined.
4	Solve \(f'(x)=0\) for \(x\) → candidate points.	Potential maxima or minima.
5	Classify each candidate using the second‑derivative test (or first‑derivative sign test).	Distinguish between maxima, minima and points of inflection.
6	Evaluate \(f(x)\) at all candidates **and** at any relevant endpoints of the domain.	Find the absolute (global) optimum.
7	Interpret the result in the context of the original problem.	State the optimum value and the condition that achieves it.

4.3 Worked Examples

Example 1 – Maximum Profit (re‑visited)

Price per unit: \(p(x)=120-0.5x\).
Weekly cost: \(C(x)=2000+20x\).
Profit \(P(x)=x\,p(x)-C(x)=100x-\tfrac12x^{2}-2000\).

1. Domain: \(x\ge0\) and price \(>0\Rightarrow x<240\).
2. \(P'(x)=100-x\). Set to zero → \(x=100\).
3. \(P''(x)=-1<0\) ⇒ local maximum.
4. Check endpoints: \(P(0)=-2000\), \(P(239)\approx-239\) (both lower).
5. Maximum profit \(P(100)=3000\) £.

Answer: Produce **100 units per week** for a maximum profit of **£3 000**.

Example 2 – Minimum Material for a Cylindrical Can

Required volume \(V=500\text{ cm}^{3}\). Surface area (material) \(S=2\pi r^{2}+2\pi r h\).

1. Constraint: \(h=\dfrac{V}{\pi r^{2}}=\dfrac{500}{\pi r^{2}}\).
2. Substitute: \(S(r)=2\pi r^{2}+2\pi r\left(\dfrac{500}{\pi r^{2}}\right)=2\pi r^{2}+\dfrac{1000}{r}\).
3. \(S'(r)=4\pi r-\dfrac{1000}{r^{2}}=0\) ⇒ \(r^{3}= \dfrac{250}{\pi}\) ⇒ \(r\approx5.42\text{ cm}\).
4. \(S''(r)=4\pi+\dfrac{2000}{r^{3}}>0\) ⇒ minimum.
5. Height \(h=\dfrac{500}{\pi r^{2}}\approx10.84\text{ cm}\).

Result: Radius ≈ 5.4 cm, height ≈ 10.8 cm gives the least material.

Example 3 – Maximum Area of a Rectangular Garden against a Wall (30 m fencing)

Let the side perpendicular to the wall be \(x\) m and the side parallel to the wall be \(y\) m. Fencing used: \(2x+y=30\) ⇒ \(y=30-2x\).
Area \(A(x)=x\,y = x(30-2x)=30x-2x^{2}\).

1. Domain: \(x>0\) and \(y>0\Rightarrow 30-2x>0\) ⇒ \(0
2. \(A'(x)=30-4x=0\) ⇒ \(x=7.5\) m.
3. \(A''(x)=-4<0\) ⇒ maximum.
4. Corresponding \(y=30-2(7.5)=15\) m.

Answer: Dimensions \(7.5\text{ m}\times15\text{ m}\) give the greatest area of \(112.5\text{ m}^{2}\).

Example 4 – Minimum Average Cost (15 marks)

Cost function: \(C(x)=0.02x^{3}-0.3x^{2}+40x+500\).
Average cost \(A(x)=\dfrac{C(x)}{x}=0.02x^{2}-0.3x+40+\dfrac{500}{x}\).

1. Domain: \(x>0\) (production level cannot be zero).
2. \(A'(x)=0.04x-0.3-\dfrac{500}{x^{2}}\). Set to zero:
   \(0.04x^{3}-0.3x^{2}-500=0\). Solve (using trial or calculator) ⇒ \(x\approx25\) units.
3. \(A''(x)=0.04+\dfrac{1000}{x^{3}}>0\) for \(x>0\) ⇒ minimum.
4. Minimum average cost \(A(25)\approx0.02(625)-0.3(25)+40+20=12.5-7.5+40+20=65\) dollars per unit.

Result: Producing **25 units** minimises the average cost to **\$65 per unit**.

Example 5 – Box with Different Material Costs (15 marks)

Square base of side \(a\) cm, height \(h\) cm, volume \(V=1000\text{ cm}^{3}\).
Base material cost = \(2c\) per cm², side material cost = \(c\) per cm² (take \(c=1\) for simplicity).

• Cost \(=2a^{2}+4ah\) (since there are four sides, each area \(a h\)).
• Constraint: \(a^{2}h=1000\) ⇒ \(h=\dfrac{1000}{a^{2}}\).
• Substitute: \(C(a)=2a^{2}+4a\left(\dfrac{1000}{a^{2}}\right)=2a^{2}+\dfrac{4000}{a}\).
• \(C'(a)=4a-\dfrac{4000}{a^{2}}=0\) ⇒ \(a^{3}=1000\) ⇒ \(a=10\) cm.
• \(C''(a)=4+\dfrac{8000}{a^{3}}>0\) ⇒ minimum.
• Height \(h=\dfrac{1000}{10^{2}}=10\) cm.

Answer: A cube of side 10 cm (base 10 cm, height 10 cm) gives the least total cost.

5. Additional Applications of Differentiation

• Rates of change: population growth, cooling of a liquid, investment value.
• Curve sketching: use \(f'(x)\) for increasing/decreasing intervals, \(f''(x)\) for concavity, points of inflection, and asymptotes.
• Implicit differentiation: differentiate equations such as \(x^{2}+y^{2}=r^{2}\) to find \(\dfrac{dy}{dx}\).
• Work and energy: \(W=\int F\,dx\), power \(P=\dfrac{dW}{dt}\).

6. Common Pitfalls & How to Avoid Them

• Ignoring domain restrictions (e.g., lengths must be positive, price cannot be negative).
• Assuming a stationary point is a maximum without a test – always apply the second‑derivative test or the first‑derivative sign test.
• Mixing up the variable of differentiation (e.g., differentiating with respect to \(t\) when the problem asks for change with respect to \(x\)).
• For optimisation, forgetting to evaluate the function at the endpoints of a closed interval.
• When integrating, omitting the constant of integration for indefinite integrals – it may be needed to satisfy an initial condition.
• In kinematics, using the wrong sign for acceleration when an object is decelerating.

7. Practice Questions (with Suggested Marks)

#	Question	Marks
1	A rectangular garden is to be built against a straight wall, using 30 m of fencing for the other three sides. Find the dimensions that give the maximum area.	6
2	The height of a ball thrown upwards is \(h(t)=20t-5t^{2}\) metres. Determine the time when the ball reaches its highest point and the maximum height.	4
3	A manufacturer’s cost function is \(C(x)=0.02x^{3}-0.3x^{2}+40x+500\) (dollars). Find the production level that minimises the average cost per unit.	8
4	For a rectangular box with a square base, the volume must be \(1000\text{ cm}^3\). The base material costs twice as much per unit area as the side material. Determine the dimensions that minimise the total cost.	10
5	Evaluate the definite integral \(\displaystyle\int_{0}^{2}\bigl(3x^{2}-4x+1\bigr)\,dx\) and interpret its meaning as the signed area between the curve and the \(x\)-axis.	4
6	Given the acceleration \(a(t)=6-2t\) m s\(^{-2}\) with initial velocity \(v(0)=4\) m s\(^{-1}\) and initial displacement \(s(0)=0\), find the expressions for \(v(t)\) and \(s(t)\) and state the total distance travelled in the first 5 s.	8
7	Find the area enclosed between the curves \(y=x^{2}\) and \(y=2x\). State which curve is the upper one on the relevant interval.	6
8	Differentiate implicitly to find \(\dfrac{dy}{dx}\) for the ellipse \(\displaystyle\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\).	5