Global maxima and minima and optimisation on restricted domains: the closed interval method, endpoints and discontinuities as candidates, domain restrictions in modelling problems and writing a contextual conclusion (local and global extrema and endpoint checks carry over from 2017; formal definitions and discontinuity checks are added in 2024)

Let the height be $h$. Surface area $2x^2 + 4xh = 600$, so $h = \frac{300 - x^2}{2x}$. Then $V = x^2 h = \frac{x(300 - x^2)}{2} = 150x - \frac{x^3}{2}$.

$V^\prime = 150 - \frac{3x^2}{2} = 0$ gives $x^2 = 100$, so $x = 10$ (the positive root). $V^{\prime\prime} = -3x < 0$, so this is a maximum. $V(10) = 1500 - 500 = 1000$ cm$^3$.

Markers award one mark for the height in terms of $x$, one for $V$, one for $x = 10$ with a reason for the root chosen and one for the volume with the nature established.

$f^\prime(x) = 6x^2 - 18x + 12 = 6(x - 1)(x - 2)$, so $x = 1$ and $x = 2$. Then $f(1) = 2 - 9 + 12 = 5$ and $f(2) = 16 - 36 + 24 = 4$. Endpoints: $f(0) = 0$ and $f(3) = 54 - 81 + 36 = 9$.

The global maximum is $9$ at $x = 3$ and the global minimum is $0$ at $x = 0$. The local maximum $5$ at $x = 1$ and the local minimum $4$ at $x = 2$ are neither. Markers reward the full list of candidates and the correct comparison.

Stationary point

$f^\prime(x) = 2x - 4 = 0$ gives $x = 2$, which lies in the interval, and $f(2) = 4 - 8 + 1 = -3$.

Endpoints

$f(0) = 1$ and $f(5) = 25 - 20 + 1 = 6$.

Compare

The values are $-3$, $1$ and $6$. The global maximum is $6$ (at $x = 5$) and the global minimum is $-3$ (at $x = 2$).

Marker's note: one mark for the stationary point, one for both endpoint values and one for the correct comparison.

Stationary points. $f^\prime(x) = 3x^2 - 3 = 3(x - 1)(x + 1) = 0$ gives $x = \pm 1$. Then $f(-1) = -1 + 3 = 2$ and $f(1) = 1 - 3 = -2$.

Endpoints. $f(-2) = -8 + 6 = -2$ and $f(3) = 27 - 9 = 18$.

Compare $2$, $-2$, $-2$ and $18$. The global maximum is $18$ at $x = 3$. The global minimum is $-2$, reached at both $x = 1$ and $x = -2$.

Marker's note: the local maximum $f(-1) = 2$ is not the global maximum. The function takes the same minimum value at two different $x$-values.

Stationary point

$f^\prime(x) = 1 - \dfrac{4}{x^2} = 0$ gives $x^2 = 4$, so $x = 2$ (the root $x = -2$ is outside the interval). $f(2) = 2 + 2 = 4$.

Endpoints

$f(1) = 1 + 4 = 5$ and $f(4) = 4 + 1 = 5$.

Compare

The global minimum is $4$ at $x = 2$. The global maximum is $5$, and it occurs at both endpoints $x = 1$ and $x = 4$.

Marker's note: one mark for the derivative and root inside the domain, one for both endpoints, one for the conclusion.

(a) Build the model. The base is a square of side $12 - 2x$ and the height is $x$, so $V = x(12 - 2x)^2$. The base side must be positive and $x$ must be positive, so the domain is $0 < x < 6$; the closed interval $0 \leq x \leq 6$ is used when checking endpoints, where the volume is $0$.

(b) Differentiate. Expanding, $V = 4x^3 - 48x^2 + 144x$, so

Stationary points are $x = 2$ and $x = 6$. Only $x = 2$ is inside the open domain.

Evaluate. $V(2) = 2 \times 8^2 = 128$. At the endpoints $V(0) = 0$ and $V(6) = 0$. So the global maximum volume is $128$ cm$^3$ when $x = 2$ cm.

Marker's note: one mark for the model, one for the domain, one for the derivative, one for rejecting $x = 6$ with a reason and one for the volume with units.

Examine each piece. On $0 \leq x < 2$, $f(x) = x^2$ is increasing, from $f(0) = 0$ and approaching $4$ as $x \to 2$ from the left. The value $4$ is not attained by this piece.

On $2 \leq x \leq 5$, $f(x) = 7 - x$ is decreasing, from $f(2) = 5$ to $f(5) = 2$.

Compare. The function is discontinuous at $x = 2$ (the left piece approaches $4$ but $f(2) = 5$). The largest value actually reached is $f(2) = 5$, so the global maximum is $5$ at $x = 2$. The smallest value is $f(0) = 0$, the global minimum at $x = 0$.

Marker's note: one mark for each piece analysed, one for recognising the jump at $x = 2$ and one for the correct extrema. There are no stationary points to find here.

(a) Model

The side parallel to the wall is $100 - 2x$, so $A = x(100 - 2x) = 100x - 2x^2$. The parallel side must satisfy $100 - 2x \leq 30$, giving $x \geq 35$, and it must be non-negative, giving $x \leq 50$. So the domain is $35 \leq x \leq 50$.

(b) Differentiate

$A^\prime = 100 - 4x$, which is zero at $x = 25$. But $x = 25$ is outside the domain, so there is no stationary point to consider.

Check the endpoints

On $35 \leq x \leq 50$, $A^\prime = 100 - 4x < 0$, so $A$ is decreasing and the maximum is at the left endpoint:

The maximum area is $1050$ m$^2$ when $x = 35$ m (so the parallel side is $30$ m).

Marker's note: one mark for the model, one for the domain, one for rejecting $x = 25$, one for the endpoint comparison and one for the contextual conclusion. Without the wall limit, the unrestricted maximum would be $A(25) = 1250$ m$^2$, which is not allowed here.