Secondary 2 Mathematics Calculus Quiz

<!-- TuitionGoWhere generation metadata: stage=5-1; model=nvidia/nemotron-3-ultra-550b-a55b:free; model_label=NVIDIA Nemotron 3 Ultra 550B A55B Free; generated=2026-06-16; Sources: Stage 4-0 LLM templates, syllabus context, and Stage 2 evidence where available. -->

Secondary 2 Mathematics Quiz - Calculus

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: _____ / 40

Duration: 45 minutes
Total Marks: 40

Instructions:

• Answer all questions.
• Write your answers in the spaces provided.
• Show all working clearly.
• Omission of essential working will result in loss of marks.
• The use of an approved scientific calculator is expected, where appropriate.
• If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place.
• For π, use either your calculator value or 3.142, unless the question requires the answer in terms of π.

---

Section A: Gradient and Rate of Change (Questions 1–5) [10 marks]

1. The graph of $y = x^2$ is drawn. Find the gradient of the tangent to the curve at the point where $x = 3$.
[2 marks]

Answer: ___________________________

2. A particle moves along a straight line such that its distance $s$ metres from a fixed point $O$ after $t$ seconds is given by $s = 2t^2 + 3t$. Find the speed of the particle when $t = 4$ seconds.
[2 marks]

Answer: ___________________________

3. The gradient of the curve $y = kx^3$ at the point where $x = 2$ is 48. Find the value of the constant $k$.
[2 marks]

Answer: ___________________________

4. The volume $V$ cm³ of a sphere is given by $V = \frac{4}{3}\pi r^3$, where $r$ cm is the radius. Find the rate of change of volume with respect to radius when $r = 5$ cm.
[2 marks]

Answer: ___________________________

5. The curve $y = x^2 - 4x + 7$ has a tangent at point $P$ with gradient 2. Find the coordinates of $P$.
[2 marks]

Answer: ___________________________

---

Section B: Differentiation of Polynomials (Questions 6–10) [10 marks]

6. Differentiate the following with respect to $x$: (a) $y = 5x^4 - 3x^2 + 7$
(b) $y = \frac{2}{x^3} + 4\sqrt{x}$
[3 marks]

Answer: ___________________________

7. Given that $y = (x + 2)(x^2 - 3x + 1)$, find $\frac{dy}{dx}$ in its simplest form.
[2 marks]

Answer: ___________________________

8. The curve $y = x^3 - 6x^2 + 9x + 2$ has two stationary points. Find the coordinates of both stationary points and determine their nature.
[3 marks]

Answer: ___________________________

9. A rectangular sheet of metal measures 20 cm by 12 cm. Equal squares of side $x$ cm are cut from each corner and the sides are folded up to form an open box. (a) Show that the volume $V$ cm³ of the box is given by $V = 4x^3 - 64x^2 + 240x$.
(b) Find the value of $x$ for which $V$ has a stationary value.
[2 marks]

Answer: ___________________________

10. The gradient function of a curve is given by $\frac{dy}{dx} = 3x^2 - 6x$. The curve passes through the point $(1, 4)$. Find the equation of the curve.
[2 marks]

Answer: ___________________________

---

Section C: Applications and Problem Solving (Questions 11–15) [10 marks]

11. A stone is thrown vertically upwards and its height $h$ metres after $t$ seconds is given by $h = 20t - 5t^2$. (a) Find the initial velocity of the stone.
(b) Find the maximum height reached by the stone.
[2 marks]

Answer: ___________________________

12. The curve $y = x^3 - 3x^2 - 9x + 5$ crosses the $x$-axis at three points. Find the coordinates of the turning points of the curve.
[2 marks]

Answer: ___________________________

13. A cylindrical tank of radius $r$ cm and height $h$ cm has a fixed volume of $500\pi$ cm³. The total surface area $A$ cm² is given by $A = 2\pi r^2 + 2\pi rh$. (a) Express $h$ in terms of $r$.
(b) Show that $A = 2\pi r^2 + \frac{1000\pi}{r}$.
[2 marks]

Answer: ___________________________

14. The curve $y = 2x^3 - 9x^2 + 12x - 3$ has two turning points. Find the coordinates of these turning points and determine their nature.
[2 marks]

Answer: ___________________________

15. A wire of length 60 cm is cut into two pieces. One piece is bent to form a circle of radius $r$ cm, and the other is bent to form a square of side $x$ cm. (a) Express $x$ in terms of $r$.
(b) Show that the total area $A$ cm² enclosed by the circle and square is given by $A = \pi r^2 + \left(15 - \frac{\pi r}{2}\right)^2$.
[2 marks]

Answer: ___________________________

---

Section D: Advanced Applications (Questions 16–20) [10 marks]

16. The curve $y = x^3 - 4x^2 + 5x - 2$ has a tangent at the point where $x = 2$. Find the equation of this tangent.
[2 marks]

Answer: ___________________________

17. A particle moves along a straight line such that its velocity $v$ m/s after $t$ seconds is given by $v = 3t^2 - 12t + 9$. The particle starts from rest at $t = 0$. (a) Find the acceleration of the particle when $t = 2$ seconds.
(b) Find the distance travelled by the particle in the first 3 seconds.
[2 marks]

Answer: ___________________________

18. The volume $V$ cm³ of a cone with radius $r$ cm and height $h$ cm is given by $V = \frac{1}{3}\pi r^2 h$. If the cone has a fixed slant height of 10 cm, express $V$ in terms of $r$ only and find the value of $r$ that maximises the volume.
[2 marks]

Answer: ___________________________

19. The curve $y = x^4 - 4x^3 + 6x^2 - 4x + 1$ has a stationary point at $x = 1$. Determine the nature of this stationary point.
[2 marks]

Answer: ___________________________

20. A rectangular garden of area 200 m² is to be fenced on three sides, with the fourth side being a wall. Find the dimensions of the garden that minimise the length of fencing required.
[2 marks]

Answer: ___________________________

---

End of Quiz

<!-- TuitionGoWhere generation metadata: stage=5-1; model=nvidia/nemotron-3-ultra-550b-a55b:free; model_label=NVIDIA Nemotron 3 Ultra 550B A55B Free; generated=2026-06-16; Sources: Stage 4-0 LLM templates, syllabus context, and Stage 2 evidence where available. -->

Secondary 2 Mathematics Quiz - Calculus (Answer Key)

Total Marks: 40

---

Section A: Gradient and Rate of Change (Questions 1–5) [10 marks]

1. [2 marks]

Question: The graph of $y = x^2$ is drawn. Find the gradient of the tangent to the curve at the point where $x = 3$.

Solution: For $y = x^2$, the gradient function (derivative) is $\frac{dy}{dx} = 2x$. At $x = 3$, gradient $= 2(3) = 6$.

Answer: 6
Marking: M1 for correct derivative $2x$, A1 for correct substitution and answer 6.

---

2. [2 marks]

Question: A particle moves along a straight line such that its distance $s$ metres from a fixed point $O$ after $t$ seconds is given by $s = 2t^2 + 3t$. Find the speed of the particle when $t = 4$ seconds.

Solution: Speed $v = \frac{ds}{dt} = \frac{d}{dt}(2t^2 + 3t) = 4t + 3$. At $t = 4$, $v = 4(4) + 3 = 16 + 3 = 19$ m/s.

Answer: 19 m/s
Marking: M1 for correct differentiation, A1 for correct substitution and answer with units.

---

3. [2 marks]

Question: The gradient of the curve $y = kx^3$ at the point where $x = 2$ is 48. Find the value of the constant $k$.

Solution: $y = kx^3 \Rightarrow \frac{dy}{dx} = 3kx^2$. At $x = 2$, gradient $= 3k(2)^2 = 12k$. Given gradient $= 48$, so $12k = 48 \Rightarrow k = 4$.

Answer: $k = 4$
Marking: M1 for correct derivative $3kx^2$, M1 for substitution and equation, A1 for $k = 4$.

---

4. [2 marks]

Question: The volume $V$ cm³ of a sphere is given by $V = \frac{4}{3}\pi r^3$, where $r$ cm is the radius. Find the rate of change of volume with respect to radius when $r = 5$ cm.

Solution: $\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2$. At $r = 5$, $\frac{dV}{dr} = 4\pi(5)^2 = 4\pi(25) = 100\pi$ cm³/cm.

Answer: $100\pi$ cm³/cm (or approximately 314 cm³/cm)
Marking: M1 for correct derivative $4\pi r^2$, A1 for correct substitution and answer.

---

5. [2 marks]

Question: The curve $y = x^2 - 4x + 7$ has a tangent at point $P$ with gradient 2. Find the coordinates of $P$.

Solution: $\frac{dy}{dx} = 2x - 4$. Set gradient $= 2$: $2x - 4 = 2 \Rightarrow 2x = 6 \Rightarrow x = 3$. When $x = 3$, $y = 3^2 - 4(3) + 7 = 9 - 12 + 7 = 4$. Coordinates of $P$ are $(3, 4)$.

Answer: $(3, 4)$
Marking: M1 for correct derivative, M1 for setting equal to 2 and solving for $x$, A1 for correct $y$-coordinate and final answer.

---

Section B: Differentiation of Polynomials (Questions 6–10) [10 marks]

6. [3 marks]

Question: Differentiate the following with respect to $x$: (a) $y = 5x^4 - 3x^2 + 7$
(b) $y = \frac{2}{x^3} + 4\sqrt{x}$

Solution: (a) $\frac{dy}{dx} = 20x^3 - 6x$ (b) Rewrite: $y = 2x^{-3} + 4x^{\frac{1}{2}}$ $\frac{dy}{dx} = -6x^{-4} + 2x^{-\frac{1}{2}} = -\frac{6}{x^4} + \frac{2}{\sqrt{x}}$

Answer: (a) $20x^3 - 6x$ (b) $-\frac{6}{x^4} + \frac{2}{\sqrt{x}}$
Marking: (a) B1 for each correct term (2 marks). (b) B1 for rewriting correctly, B1 for each correct differentiated term (2 marks). Total 3 marks.

---

7. [2 marks]

Question: Given that $y = (x + 2)(x^2 - 3x + 1)$, find $\frac{dy}{dx}$ in its simplest form.

Solution: Method 1 (Expand first): $y = (x + 2)(x^2 - 3x + 1) = x^3 - 3x^2 + x + 2x^2 - 6x + 2 = x^3 - x^2 - 5x + 2$ $\frac{dy}{dx} = 3x^2 - 2x - 5$

Method 2 (Product rule): $\frac{dy}{dx} = (1)(x^2 - 3x + 1) + (x + 2)(2x - 3)$ $= x^2 - 3x + 1 + 2x^2 - 3x + 4x - 6$ $= 3x^2 - 2x - 5$

Answer: $3x^2 - 2x - 5$
Marking: M1 for correct expansion or product rule application, A1 for correct simplified derivative.

---

8. [3 marks]

Question: The curve $y = x^3 - 6x^2 + 9x + 2$ has two stationary points. Find the coordinates of both stationary points and determine their nature.

Solution: $\frac{dy}{dx} = 3x^2 - 12x + 9$ At stationary points, $\frac{dy}{dx} = 0$: $3x^2 - 12x + 9 = 0 \Rightarrow x^2 - 4x + 3 = 0 \Rightarrow (x - 1)(x - 3) = 0$ $x = 1$ or $x = 3$

When $x = 1$: $y = 1 - 6 + 9 + 2 = 6$ → Point $(1, 6)$ When $x = 3$: $y = 27 - 54 + 27 + 2 = 2$ → Point $(3, 2)$

Nature (Second derivative test): $\frac{d^2y}{dx^2} = 6x - 12$ At $x = 1$: $\frac{d^2y}{dx^2} = 6(1) - 12 = -6 < 0$ → Maximum at $(1, 6)$ At $x = 3$: $\frac{d^2y}{dx^2} = 6(3) - 12 = 6 > 0$ → Minimum at $(3, 2)$

Answer: Maximum at $(1, 6)$, Minimum at $(3, 2)$
Marking: M1 for correct first derivative, M1 for setting to zero and solving, A1 for both $x$-values, A1 for both $y$-coordinates, A1 for correct nature determination.

---

9. [2 marks]

Question: A rectangular sheet of metal measures 20 cm by 12 cm. Equal squares of side $x$ cm are cut from each corner and the sides are folded up to form an open box. (a) Show that the volume $V$ cm³ of the box is given by $V = 4x^3 - 64x^2 + 240x$.
(b) Find the value of $x$ for which $V$ has a stationary value.

Solution: (a) After cutting squares of side $x$: Length $= 20 - 2x$, Width $= 12 - 2x$, Height $Height$Height = x$$V = x(20 - 2x)(12 - 2x) = x(240 - 40x - 24x + 4x^2) = x(4x^2 - 64x + 240) = 4x^3 - 64x^2 + 240x$ ✓

(b) $\frac{dV}{dx} = 12x^2 - 128x + 240$ For stationary value, $\frac{dV}{dx} = 0$: $12x^2 - 128x + 240 = 0 \Rightarrow 3x^2 - 32x + 60 = 0$ $(3x - 2)(x - 30) = 0 \Rightarrow x = \frac{2}{3}$ or $x = 30$ Since $x < 6$ (half of 12 cm width), $x = \frac{2}{3}$ cm.

Answer: (a) Shown. (b) $x = \frac{2}{3}$ cm.
Marking: (a) M1 for correct dimensions, M1 for correct expansion, A1 for shown result. (b) M1 for correct derivative, M1 for setting to zero, M1 for solving quadratic, A1 for selecting valid root $x = \frac{2}{3}$.

---

10. [2 marks]

Question: The gradient function of a curve is given by $\frac{dy}{dx} = 3x^2 - 6x$. The curve passes through the point $(1, 4)$. Find the equation of the curve.

Solution: $y = \int (3x^2 - 6x) \, dx = x^3 - 3x^2 + c$ Substitute $(1, 4)$: $4 = 1^3 - 3(1)^2 + c = 1 - 3 + c = -2 + c$ $c = 6$ Equation: $y = x^3 - 3x^2 + 6$

Answer: $y = x^3 - 3x^2 + 6$
Marking: M1 for correct integration, M1 for substituting point to find $c$, A1 for correct final equation.

---

Section C: Applications and Problem Solving (Questions 11–15) [10 marks]

11. [2 marks]

Question: A stone is thrown vertically upwards and its height $h$ metres after $t$ seconds is given by $h = 20t - 5t^2$. (a) Find the initial velocity of the stone.
(b) Find the maximum height reached by the stone.

Solution: (a) Velocity $v = \frac{dh}{dt} = 20 - 10t$. Initial velocity at $t = 0$: $v = 20$ m/s.

(b) At maximum height, $v = 0$: $20 - 10t = 0 \Rightarrow t = 2$ s. Maximum height $h = 20(2) - 5(2)^2 = 40 - 20 = 20$ m.

Answer: (a) 20 m/s. (b) 20 m.
Marking: (a) M1 for derivative, A1 for 20 m/s. (b) M1 for setting $v=0$, M1 for finding $t=2$, A1 for $h=20$.

---

12. [2 marks]

Question: The curve $y = x^3 - 3x^2 - 9x + 5$ crosses the $x$-axis at three points. Find the coordinates of the turning points of the curve.

Solution: $\frac{dy}{dx} = 3x^2 - 6x - 9$ At turning points, $\frac{dy}{dx} = 0$: $3x^2 - 6x - 9 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0$ $x = 3$ or $x = -1$

When $x = 3$: $y = 27 - 27 - 27 + 5 = -22$ → $(3, -22)$ When $x = -1$: $y = -1 - 3 + 9 + 5 = 10$ → $(-1, 10)$

Answer: $(3, -22)$ and $(-1, 10)$
Marking: M1 for correct derivative, M1 for setting to zero and solving, A1 for both $x$-values, A1 for both $y$-coordinates.

---

13. [2 marks]

Question: A cylindrical tank of radius $r$ cm and height $h$ cm has a fixed volume of $500\pi$ cm³. The total surface area $A$ cm² is given by $A = 2\pi r^2 + 2\pi rh$. (a) Express $h$ in terms of $r$.
(b) Show that $A = 2\pi r^2 + \frac{1000\pi}{r}$.

Solution: (a) Volume $V = \pi r^2 h = 500\pi \Rightarrow r^2 h = 500 \Rightarrow h = \frac{500}{r^2}$

(b) $A = 2\pi r^2 + 2\pi r\left(\frac{500}{r^2}\right) = 2\pi r^2 + \frac{1000\pi}{r}$ ✓

Answer: (a) $h = \frac{500}{r^2}$ (b) Shown.
Marking: (a) M1 for using volume formula, A1 for correct expression. (b) M1 for substitution, A1 for correct simplification.

---

14. [2 marks]

Question: The curve $y = 2x^3 - 9x^2 + 12x - 3$ has two turning points. Find the coordinates of these turning points and determine their nature.

Solution: $\frac{dy}{dx} = 6x^2 - 18x + 12$ At turning points, $\frac{dy}{dx} = 0$: $6x^2 - 18x + 12 = 0 \Rightarrow x^2 - 3x + 2 = 0 \Rightarrow (x - 1)(x - 2) = 0$ $x = 1$ or $x = 2$

When $x = 1$: $y = 2 - 9 + 12 - 3 = 2$ → $(1, 2)$ When $x = 2$: $y = 16 - 36 + 24 - 3 = 1$ → $(2, 1)$

Nature (Second derivative test): $\frac{d^2y}{dx^2} = 12x - 18$ At $x = 1$: $\frac{d^2y}{dx^2} = 12(1) - 18 = -6 < 0$ → Maximum at $(1, 2)$ At $x = 2$: $\frac{d^2y}{dx^2} = 12(2) - 18 = 6 > 0$ → Minimum at $(2, 1)$

Answer: Maximum at $(1, 2)$, Minimum at $(2, 1)$
Marking: M1 for correct first derivative, M1 for setting to zero and solving, A1 for both $x$-values, A1 for both $y$-coordinates, A1 for correct nature determination.

---

15. [2 marks]

Question: A wire of length 60 cm is cut into two pieces. One piece is bent to form a circle of radius $r$ cm, and the other is bent to form a square of side $x$ cm. (a) Express $x$ in terms of $r$.
(b) Show that the total area $A$ cm² enclosed by the circle and square is given by $A = \pi r^2 + \left(15 - \frac{\pi r}{2}\right)^2$.

Solution: (a) Perimeter of circle $= 2\pi r$, Perimeter of square $= 4x$. Total length: $2\pi r + 4x = 60 \Rightarrow 4x = 60 - 2\pi r \Rightarrow x = 15 - \frac{\pi r}{2}$

(b) Area of circle $= \pi r^2$, Area of square $= x^2 = \left(15 - \frac{\pi r}{2}\right)^2$. Total area $A = \pi r^2 + \left(15 - \frac{\pi r}{2}\right)^2$ ✓

Answer: (a) $x = 15 - \frac{\pi r}{2}$ (b) Shown.
Marking: (a) M1 for correct perimeter equation, A1 for correct expression. (b) M1 for area expressions, A1 for correct substitution and result.

---

Section D: Advanced Applications (Questions 16–20) [10 marks]

16. [2 marks]

Question: The curve $y = x^3 - 4x^2 + 5x - 2$ has a tangent at the point where $x = 2$. Find the equation of this tangent.

Solution: $\frac{dy}{dx} = 3x^2 - 8x + 5$ At $x = 2$, gradient $m = 3(4) - 8(2) + 5 = 12 - 16 + 5 = 1$ When $x = 2$, $y = 8 - 16 + 10 - 2 = 0$ → Point $(2, 0)$ Equation of tangent: $y - 0 = 1(x - 2) \Rightarrow y = x - 2$

Answer: $y = x - 2$
Marking: M1 for correct derivative, M1 for finding gradient at $x=2$, M1 for finding point on curve, A1 for correct tangent equation.

---

17. [2 marks]

Question: A particle moves along a straight line such that its velocity $v$ m/s after $t$ seconds is given by $v = 3t^2 - 12t + 9$. The particle starts from rest at $t = 0$. (a) Find the acceleration of the particle when $t = 2$ seconds.
(b) Find the distance travelled by the particle in the first 3 seconds.

Solution: (a) Acceleration $a = \frac{dv}{dt} = 6t - 12$ At $t = 2$, $a = 6(2) - 12 = 0$ m/s²

(b) Distance $s = \int v \, dt = \int (3t^2 - 12t + 9) \, dt = t^3 - 6t^2 + 9t + c$ Since particle starts from rest at $t=0$, we assume $s=0$ at $t=0$, so $c=0$. Distance in first 3 seconds: $s(3) - s(0) = (27 - 54 + 27) - 0 = 0$ m (Note: The particle returns to its starting point at $t=3$)

Answer: (a) 0 m/s² (b) 0 m
Marking: (a) M1 for correct derivative, A1 for correct value. (b) M1 for correct integration, M1 for limits/constant, A1 for correct evaluation.

---

18. [2 marks]

Question: The volume $V$ cm³ of a cone with radius $r$ cm and height $h$ cm is given by $V = \frac{1}{3}\pi r^2 h$. If the cone has a fixed slant height of 10 cm, express $V$ in terms of $r$ only and find the value of $r$ that maximises the volume.

Solution: Slant height $l = 10$, so $l^2 = r^2 + h^2 \Rightarrow 100 = r^2 + h^2 \Rightarrow h = \sqrt{100 - r^2}$ $V = \frac{1}{3}\pi r^2 \sqrt{100 - r^2}$ To maximise $V$, maximise $V^2$ (or use derivative directly): Let $u = V^2 = \frac{1}{9}\pi^2 r^4 (100 - r^2) = \frac{1}{9}\pi^2 (100r^4 - r^6)$ $\frac{du}{dr} = \frac{1}{9}\pi^2 (400r^3 - 6r^5) = 0 \Rightarrow 2r^3(200 - 3r^2) = 0$ $r = 0$ (minimum) or $r^2 = \frac{200}{3} \Rightarrow r = \sqrt{\frac{200}{3}} = \frac{10\sqrt{6}}{3}$ cm

Answer: $V = \frac{1}{3}\pi r^2 \sqrt{100 - r^2}$, $r = \frac{10\sqrt{6}}{3}$ cm (or approximately 8.16 cm)
Marking: M1 for expressing $h$ in terms of $r$, M1 for $V$ in terms of $r$, M1 for differentiation, M1 for solving, A1 for correct $r$ value.

---

19. [2 marks]

Question: The curve $y = x^4 - 4x^3 + 6x^2 - 4x + 1$ has a stationary point at $x = 1$. Determine the nature of this stationary point.

Solution: $\frac{dy}{dx} = 4x^3 - 12x^2 + 12x - 4 = 4(x^3 - 3x^2 + 3x - 1) = 4(x - 1)^3$ At $x = 1$, $\frac{dy}{dx} = 0$ ✓ (stationary point)

$\frac{d^2y}{dx^2} = 12x^2 - 24x + 12 = 12(x^2 - 2x + 1) = 12(x - 1)^2$ At $x = 1$, $\frac{d^2y}{dx^2} = 0$ (inconclusive)

Use first derivative test: For $x < 1$ (e.g., $x = 0$): $\frac{dy}{dx} = 4(-1)^3 = -4 < 0$ For $x > 1$ (e.g., $x = 2$): $\frac{dy}{dx} = 4(1)^3 = 4 > 0$ Since derivative changes from negative to positive, $x = 1$ is a minimum point.

Alternatively, note $y = (x - 1)^4 \ge 0$, with equality at $x = 1$, so minimum.

Answer: Minimum point at $(1, 0)$
Marking: M1 for first derivative, M1 for second derivative (showing 0), M1 for first derivative test or alternative reasoning, A1 for correct conclusion.

---

20. [2 marks]

Question: A rectangular garden of area 200 m² is to be fenced on three sides, with the fourth side being a wall. Find the dimensions of the garden that minimise the length of fencing required.

Solution: Let length parallel to wall $= x$ m, width perpendicular to wall $= y$ m. Area: $xy = 200 \Rightarrow y = \frac{200}{x}$ Fencing length $L = x + 2y = x + \frac{400}{x}$ $\frac{dL}{dx} = 1 - \frac{400}{x^2}$ For minimum, $\frac{dL}{dx} = 0 \Rightarrow 1 = \frac{400}{x^2} \Rightarrow x^2 = 400 \Rightarrow x = 20$ m (positive) Then $y = \frac{200}{20} = 10$ m Second derivative: $\frac{d^2L}{dx^2} = \frac{800}{x^3} > 0$ for $x > 0$, so minimum.

Answer: Length = 20 m, Width = 10 m (or 20 m parallel to wall, 10 m perpendicular)
Marking: M1 for defining variables and area constraint, M1 for fencing length expression, M1 for differentiation and setting to zero, A1 for correct dimensions, A1 for verification of minimum.

---

End of Answer Key