Stationary Points & Classification Tests
A stationary point occurs wherever the instantaneous gradient is zero:
To classify whether a stationary point is a local maximum or local minimum:
The Second Derivative Test:
- If $f''(a) < 0 \implies$ Curve is concave down ($\cap$), so $(a, f(a))$ is a local maximum.
- If $f''(a) > 0 \implies$ Curve is concave up ($\cup$), so $(a, f(a))$ is a local minimum.
- If $f''(a) = 0 \implies$ The test is inconclusive; use the First Derivative Sign Test.
Step-by-Step Applied Optimization Protocol
- Identify the Objective: Write a formula for the quantity to be maximized or minimized (e.g. Volume $V$, Cost $C$, Profit $P$, Area $A$).
- Use the Constraint Equation: If the objective has two variables, use the given constraint (e.g. fixed perimeter or fixed volume) to express one variable in terms of the other.
- Express as a Single Variable Function: Substitute the constraint into the objective function: $f(x)$.
- Differentiate & Set to Zero: Find $f'(x)$ and solve $f'(x) = 0$.
- Verify Nature of Extrema: Compute $f''(x)$ to rigorously justify that it is indeed a maximum or minimum as required by the IB markscheme.
- Answer the Exact Question: Check whether the exam asks for the dimensions ($x$), the optimum value ($f(x)$), or both!
Classic Exam Models: Boxes, Cans & Profit
- Open Box from Cardboard: A sheet of dimensions $L \times W$ has square corners of side $x$ cut out:
$$V(x) = x(L - 2x)(W - 2x) \quad \text{with domain } 0 < x < \frac{\min(L, W)}{2}$$
- Cylindrical Can with Fixed Volume $V_0$:
$$V_0 = \pi r^2 h \implies h = \frac{V_0}{\pi r^2} \implies A(r) = 2\pi r^2 + 2\pi r h = 2\pi r^2 + \frac{2V_0}{r}$$
- Economics / Business: Profit $P(x) = R(x) - C(x)$. Maximum profit occurs where $P'(x) = 0 \iff R'(x) = C'(x)$ (Marginal Revenue = Marginal Cost).
Global extrema on a constrained domain
- On a closed bounded interval, compare all feasible stationary points, endpoints and points where the derivative does not exist. A local second-derivative test alone does not establish a global optimum.
- For an open domain the best value may be approached but never attained. State the feasible dimensions before differentiating. Discrete outputs require comparing nearby allowed integers.
- Example profit P(q)=−q²+12q−20 for0≤q≤4: unconstrained vertexq6is infeasible. P′>0on the permitted interval, so maximumP4=12.
Find the coordinates of the stationary point of $f(x) = x^2 - 8x + 11$, and use the second derivative test to justify its nature.
Solution:
$f'(x) = 2x - 8$. Set $f'(x) = 0 \implies 2x = 8 \implies x = 4$. [M1][A1]
$f(4) = 4^2 - 8(4) + 11 = 16 - 32 + 11 = -5 \implies (4, -5)$. [A1]
$f''(x) = 2$. Since $f''(4) = 2 > 0$, the stationary point is a local minimum. [R1]
An open-topped box is made from a square piece of tin of side $24\text{ cm}$ by cutting a square of side $x\text{ cm}$ from each corner and turning up the sides.
- Show that the volume of the box is given by $V(x) = 4x^3 - 96x^2 + 576x$.
- Find the value of $x$ that maximizes the volume, and find the maximum volume.
Solution:
(a) Height $= x$, Base side $= 24 - 2x$. [M1]
$V(x) = x(24 - 2x)^2 = x(576 - 96x + 4x^2) = 4x^3 - 96x^2 + 576x$. [A1]
(b) $V'(x) = 12x^2 - 192x + 576 = 0 \implies x^2 - 16x + 48 = 0$. [M1]
$(x - 4)(x - 12) = 0$. Since $0 < x < 12$, $x = 4\text{ cm}$. [A1]
Max volume: $V(4) = 4(24 - 8)^2 = 4(16^2) = 4(256) = 1024\text{ cm}^3$. [A1]
A rectangular garden is to be fenced using $80\text{ meters}$ of fencing. Find the dimensions that enclose the maximum possible area.
Solution:
Perimeter: $2x + 2y = 80 \implies x + y = 40 \implies y = 40 - x$. [M1]
Area $A(x) = x(40 - x) = 40x - x^2$. [A1]
$A'(x) = 40 - 2x = 0 \implies 2x = 40 \implies x = 20\text{ meters}$. [M1]
$y = 40 - 20 = 20\text{ meters}$. Dimensions: a $20\text{ m} \times 20\text{ m}$ square (Max area $= 400\text{ m}^2$). [A1]
A closed cylindrical can must hold a volume of $500\text{ cm}^3$. The surface area is given by $A(r) = 2\pi r^2 + \frac{1000}{r}$.
- Find $\frac{dA}{dr}$.
- Calculate the radius $r$ that minimizes the surface area of the can.
Solution:
(a) $A(r) = 2\pi r^2 + 1000r^{-1} \implies \frac{dA}{dr} = 4\pi r - 1000r^{-2} = 4\pi r - \frac{1000}{r^2}$. [M1][A1]
(b) Set $\frac{dA}{dr} = 0 \implies 4\pi r = \frac{1000}{r^2} \implies 4\pi r^3 = 1000$. [M1]
$r^3 = \frac{1000}{4\pi} = \frac{250}{\pi} \approx 79.577...$ [M1]
$r = \sqrt[3]{79.577} \approx 4.30\text{ cm}$ (to 3 s.f.). [A1]
For the cylindrical can in Exercise 4, verify using the second derivative that $r \approx 4.30\text{ cm}$ gives a minimum surface area.
Solution:
$\frac{d^2 A}{dr^2} = \frac{d}{dr}(4\pi r - 1000r^{-2}) = 4\pi + 2000r^{-3} = 4\pi + \frac{2000}{r^3}$. [M1][A1]
For $r = 4.30 > 0$, $\frac{d^2 A}{dr^2} = 4\pi + \frac{2000}{79.58} = 4\pi + 25.13 > 0$. [R1]
Since the second derivative is positive, the surface area is minimized.
A firm's daily profit is modeled by $P(x) = -0.5x^3 + 15x^2 + 100x - 200$, where $x$ is production output in units ($0 \le x \le 30$). Find the production level that maximizes profit.
Solution:
$P'(x) = -1.5x^2 + 30x + 100$. [M1]
Set $P'(x) = 0 \implies -1.5x^2 + 30x + 100 = 0 \implies 3x^2 - 60x - 200 = 0$. [M1]
Using quadratic formula: $x = \frac{60 \pm \sqrt{3600 - 4(3)(-200)}}{6} = \frac{60 \pm \sqrt{6000}}{6}$. [M1]
$x = \frac{60 + 77.46}{6} = 22.91\text{ units}$ (reject negative root). Profit is maximized at $x \approx 23\text{ units}$. [A1]
Explain why checking domain endpoints is essential when finding the global maximum of a function defined on a closed interval $[a, b]$.
Solution:
Setting $f'(x) = 0$ only identifies interior stationary points (local extrema). [R1]
On a restricted domain $[a, b]$, the absolute highest or lowest value can occur at the boundary endpoints $x = a$ or $x = b$, even if $f'(x) \ne 0$ there. [A2]
Find the stationary points of $f(x) = 2x^3 - 9x^2 + 12x + 1$ and classify each point.
Solution:
$f'(x) = 6x^2 - 18x + 12 = 0 \implies x^2 - 3x + 2 = 0 \implies (x - 1)(x - 2) = 0$. [M1][A1]
Stationary points at $x = 1$ and $x = 2$.
$f''(x) = 12x - 18$.
At $x = 1$: $f''(1) = 12(1) - 18 = -6 < 0 \implies$ Local maximum at $(1, 6)$. [A1]
At $x = 2$: $f''(2) = 12(2) - 18 = +6 > 0 \implies$ Local minimum at $(2, 5)$. [A1]
A window consists of a rectangle surmounted by a semicircle. The perimeter of the window is $6\text{ meters}$. If the radius of the semicircle is $r$, express the area $A$ in terms of $r$ alone.
Solution:
Width $= 2r$, height of rectangle $= h$. Semicircle perimeter $= \pi r$.
Total perimeter: $2r + 2h + \pi r = 6 \implies 2h = 6 - 2r - \pi r \implies h = 3 - r - \frac{\pi}{2}r$. [M1][A1]
Area: $A = (2r)h + \frac{1}{2}\pi r^2 = 2r\left(3 - r - \frac{\pi}{2}r\right) + \frac{1}{2}\pi r^2$. [M1]
$A(r) = 6r - 2r^2 - \pi r^2 + 0.5\pi r^2 = 6r - \left(2 + \frac{\pi}{2}\right)r^2$. [A1]
Using the area formula from Exercise 9, $A(r) = 6r - \left(2 + \frac{\pi}{2}\right)r^2$, calculate the value of $r$ that maximizes light transmission through the window.
Solution:
$A'(r) = 6 - 2\left(2 + \frac{\pi}{2}\right)r = 6 - (4 + \pi)r$. [M1]
Set $A'(r) = 0 \implies (4 + \pi)r = 6$. [M1]
$r = \frac{6}{4 + \pi} \approx \frac{6}{7.14159} \approx 0.840\text{ meters}$. [A1]
$A''(r) = -(4 + \pi) < 0 \implies$ Confirms maximum area. [R1]
Find the global minimum and maximum of $f(x)=x^3-3x$ on[-2,3].
f′=3x²−3givesx±1. Candidate valuesf−2=−2,f−1=2,f1=−2,f3=18. Global minimum−2at−2and1; maximum18at3.
Profit is $P(n)=-(n-5.4)^2+30$ for integer nfrom0to10. Find the best output.
Continuous vertex5.4; compare integers5and6:P5=29.84,P6=29.64. Bestn5. Simply rounding a reported optimum without comparison is insufficient.
Core Formulas (Topic 5.4 - 5.6)
| Stationary Point Condition | $$f'(x) = 0$$ |
| Second Derivative Test: Maximum | $$f'(a) = 0 \quad \text{and} \quad f''(a) < 0$$ |
| Second Derivative Test: Minimum | $$f'(a) = 0 \quad \text{and} \quad f''(a) > 0$$ |
| Profit Maximization | $$P(x) = R(x) - C(x) \implies P'(x) = 0 \iff R'(x) = C'(x)$$ |
Top 5 IB Exam Traps to Avoid
- Trap 1: Forgetting to test the nature: Finding $x$ from $f'(x) = 0$ is only half the problem. You must state $f''(x)$ or show a first derivative sign table to secure the reasoning marks.
- Trap 2: Confusing the maximum value with the location: If asked for the "maximum area", evaluate $A(x_{\max})$. Do not stop at $x_{\max}$.
- Trap 3: Constraint substitution errors: Always double check that the objective function is expressed purely in terms of a single variable before differentiating.
- Trap 4: Inverting the second derivative test: Remember: $f''(x) < 0$ means maximum (frowning curve $\cap$); $f''(x) > 0$ means minimum (smiling curve $\cup$).
- Trap 5: Discarding domain limits: In real problems (e.g. cut box corners), verify that your stationary point lies within the physical domain ($0 < x < L/2$).
5-Question Concept Check
Score: - / 51. If $f'(3) = 0$ and $f''(3) = -8$, what type of stationary point is at $x = 3$?
2. At which $x$-value does $f(x) = 2x^3 - 6x$ have a local minimum?
3. For an open box made by cutting corners of size $x$ from a $10 \times 16\text{ cm}$ card, what is the domain of $x$?
4. In economics, maximum profit occurs when:
5. If $f''(x) = 0$ at a stationary point, what should you do?