Use the first and second derivatives to find stationary points, points of inflection, and to solve optimisation and related rates problems

Let $x$ be the side perpendicular to the river and $y$ the side parallel to it.

Constraint: $2x + y = 200$, so $y = 200 - 2x$.

Area: $A(x) = x y = x(200 - 2x) = 200 x - 2 x^2$.

Differentiate: $A'(x) = 200 - 4x$. Set $A'(x) = 0$: $x = 50$.

Second derivative: $A''(x) = -4 < 0$, so this is a maximum.

$y = 200 - 100 = 100$. Maximum area $= 50 \times 100 = 5000$ m$^2$.

Markers reward defining variables, writing the constraint, expressing area in one variable, and confirming a maximum using $A''$ or sign analysis.

By similar triangles, $\frac{r}{h} = \frac{6}{12} = \frac{1}{2}$, so $r = \frac{h}{2}$.

Volume of water: $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}$.

Differentiate with respect to $t$: $\frac{dV}{dt} = \frac{\pi h^2}{4} \cdot \frac{dh}{dt}$.

Substitute $\frac{dV}{dt} = 4$ and $h = 3$: $4 = \frac{9 \pi}{4} \cdot \frac{dh}{dt}$.

$\frac{dh}{dt} = \frac{16}{9 \pi} \approx 0.566$ cm/s.

Markers expect the similar-triangle relation, volume in one variable, implicit differentiation in $t$, and a final answer with units.

Differentiate and set the gradient to zero. A stationary point occurs where $\frac{dy}{dx} = 0$:

Find the $y$-coordinate. Substitute $x = 4$ back into the curve:

so the stationary point is $(4, -11)$.

Classify it with the second derivative. Here $\frac{d^2y}{dx^2} = 2 > 0$, so the curve is concave up and the point is a minimum.

Marker's note: one mark for solving $\frac{dy}{dx} = 0$ to get $x = 4$; one mark for the full coordinates $(4, -11)$ with the nature justified (by $y'' > 0$ or the fact that the coefficient of $x^2$ is positive). Stating "minimum" without any justification does not earn the second mark.

Locate the stationary points. Solve $f'(x) = 0$:

Find both $y$-coordinates.

giving the points $(1, -2)$ and $(-1, 2)$.

Classify with $f''(x) = 6x$.

Marker's note: one mark for both $x$-values from $f'(x) = 0$, one for the two coordinate pairs, one for classifying each correctly with $f''$. A candidate who finds $x = \pm 1$ but never classifies loses the final mark.

Find the point of contact. At $x = 1$, $y = 1^2 - 4(1) = -3$, so the tangent touches at $(1, -3)$.

Find the gradient there. Differentiate and substitute:

Write the line through $(1, -3)$ with gradient $-2$.

Marker's note: one mark for the point $(1, -3)$, one for the gradient $f'(1) = -2$, one for a correctly arranged line equation. Using the gradient formula but forgetting to evaluate at $x = 1$ (leaving $2x - 4$) is the usual slip.

Stationary points: solve $y' = 0$.

so the stationary points are $(1, 4)$ and $(3, 0)$.

Classify with $y'' = 6x - 12$.

Point of inflection: solve $y'' = 0$. $6x - 12 = 0$ gives $x = 2$, and $y''$ changes sign from negative to positive there, so the concavity genuinely changes. With $y(2) = 8 - 24 + 18 = 2$, the inflection is at $(2, 2)$.

Marker's note: one mark for the stationary $x$-values, one for their coordinates, one for classifying both with $y''$, one for the inflection point confirmed by a change in concavity (not merely $y'' = 0$).

Read everything from the sign of $f'(x)$, because $f'$ is what the graph shows.

Part (a): stationary points sit where $f'(x) = 0$. The gradient graph crosses the axis at $x = -2$ and $x = 4$, so $f(x)$ has stationary points at $x = -2$ and $x = 4$.

Part (b): classify by how $f'$ changes sign. For $x < -2$ the parabola is above the axis, so $f'(x) > 0$ (rising); for $-2 < x < 4$ it is below, so $f'(x) < 0$ (falling); for $x > 4$ it is above again, so $f'(x) > 0$.

• At $x = -2$, $f'$ changes from $+$ to $-$, so $f$ turns from increasing to decreasing: a local maximum.
• At $x = 4$, $f'$ changes from $-$ to $+$, so $f$ turns from decreasing to increasing: a local minimum.

Part (c): $f$ decreases where $f'(x) < 0$. That is exactly where the graph is below the axis, so $f(x)$ is decreasing on $-2 < x < 4$.

Marker's note: the trap is treating the plotted parabola as $f(x)$ itself. It is $f'(x)$, so its $x$-intercepts are the stationary points of $f$ and its sign gives increasing/decreasing. One mark each for (a) and (c); two for classifying both stationary points by the sign change in part (b).

Write the link between the two changing quantities. The area of a circle of radius $r$ is

Differentiate with respect to time. Since both $A$ and $r$ change with $t$, differentiate implicitly:

Substitute the instant. We are given $\frac{dr}{dt} = 2$ and want the rate when $r = 10$:

Marker's note: one mark for $A = \pi r^2$ and the chain-rule link $\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$, one for correct substitution of $r = 10$ and $\frac{dr}{dt} = 2$, one for the answer $40\pi$ with units. Differentiating $A$ with respect to $r$ only (forgetting the $\frac{dr}{dt}$ factor) is the standard error.

Part (a): build the volume. After cutting squares of side $x$ from each corner and folding up, the base is a square of side $24 - 2x$ and the height is $x$. So

valid for $0 < x < 12$ (the flaps must have positive height and the base positive side).

Part (b): differentiate using the product and chain rules.

Factor out $(24 - 2x)$:

Set $V' = 0$: either $24 - 2x = 0$ (so $x = 12$, which gives $V = 0$ and is rejected) or $24 - 6x = 0$, giving $x = 4$.

Justify the maximum with a sign test. For $x$ just below $4$, both factors are positive, so $V' > 0$; for $x$ just above $4$, $(24 - 6x) < 0$ while $(24 - 2x) > 0$, so $V' < 0$. The gradient changes $+$ to $-$, so $x = 4$ gives a maximum.

Part (c): evaluate.

Marker's note: one mark for showing $V = x(24 - 2x)^2$ with a stated domain, two for differentiating and solving to $x = 4$ (correctly rejecting $x = 12$), one for the justification that it is a maximum (sign table or $V''$), one for the final $1024$ cm$^3$ with units. Omitting the justification caps the answer even with the right $x$.

Part (a): eliminate the height using the fixed volume. An open-top cylinder has one circular base and a curved side:

The volume fixes $h$:

Substitute to remove $h$:

Part (b): differentiate and solve $S' = 0$.

Justify the minimum with the second derivative.

so the curve is concave up and $r = 5$ gives the minimum surface area.

Part (c): find the height.

Marker's note: one mark for $S = \pi r^2 + 2\pi r h$ with the volume substitution, one for reaching the shown form, one for solving $S' = 0$ to $r = 5$, one for the justification via $S'' > 0$, one for $h = 5$ cm. A closed-cylinder formula ($2\pi r^2 + 2\pi r h$) is a common misread of "open-top" and loses the first mark.