Write down an expression for $T$ in terms of $r$, $l$ and $\pi$.

The volume of the lobster trap is $0.75{\text{ }}{{\text{m}}^{\text{3}}}$.

Write down an equation for the volume of the lobster trap in terms of $r$, $l$ and $\pi$.

The volume of the lobster trap is $0.75{\text{ }}{{\text{m}}^{\text{3}}}$.

Show that $T = (2\pi  + 4)r + \frac{6}{{\pi {r^2}}}$.

The volume of the lobster trap is $0.75{\text{ }}{{\text{m}}^{\text{3}}}$.

Find $\frac{{{\text{d}}T}}{{{\text{d}}r}}$.

The lobster trap is designed so that the length of steel used in its frame is a minimum.

Show that the value of $r$ for which $T$ is a minimum is $0.719 {\text{ m}}$, correct to three significant figures.

The lobster trap is designed so that the length of steel used in its frame is a minimum.

Calculate the value of $l$ for which $T$ is a minimum.

The lobster trap is designed so that the length of steel used in its frame is a minimum.

Calculate the minimum value of $T$.

$2\pi r + 4r + 4l$     (A1)(A1)(A1)

Notes: Award (A1) for $2\pi r$ (“$\pi$” must be seen), (A1) for $4r$, (A1) for $4l$. Accept equivalent forms. Accept $T = 2\pi r + 4r + 4l$. Award a maximum of (A1)(A1)(A0) if extra terms are seen.

[3 marks]

$0.75 = \frac{{\pi {r^2}l}}{2}$     (A1)(A1)(A1)

Notes:     Award (A1) for their formula equated to $0.75$, (A1) for $l$ substituted into volume of cylinder formula, (A1) for volume of cylinder formula divided by $2$.

If “$\pi$” not seen in part (a) accept use of $3.14$ or greater accuracy. Award a maximum of (A1)(A1)(A0) if extra terms are seen.

[3 marks]

$T = 2\pi r + 4r + 4\left( {\frac{{1.5}}{{\pi {r^2}}}} \right)$     (A1)(ft)(A1)

$= (2\pi  + 4)r + \frac{6}{{\pi {r^2}}}$     (AG)

Notes: Award (A1)(ft) for correct rearrangement of their volume formula in part (b) seen, award (A1) for the correct substituted formula for $T$. The final line must be seen, with no incorrect working, for this second (A1) to be awarded.

[2 marks]

$\frac{{{\text{d}}T}}{{{\text{d}}r}} = 2\pi  + 4 - \frac{{12}}{{\pi {r^3}}}$     (A1)(A1)(A1)

Note: Award (A1) for $2\pi  + 4$, (A1) for $\frac{{ - 12}}{\pi }$, (A1) for ${r^{ - 3}}$.

     Accept 10.3 (10.2832…) for $2\pi  + 4$,  accept $–3.82$ $–3.81971…$ for $\frac{{ - 12}}{\pi }$. Award a maximum of (A1)(A1)(A0) if extra terms are seen.

[3 marks]

$2\pi  + 4 - \frac{{12}}{{\pi {r^3}}} = 0$   OR   $\frac{{{\text{d}}T}}{{{\text{d}}r}} = 0$     (M1)

Note: Award (M1) for setting their derivative equal to zero.

$r = 0.718843 \ldots$   OR   $\sqrt[3]{{0.371452 \ldots }}$   OR   $\sqrt[3]{{\frac{{12}}{{\pi (2\pi  + 4)}}}}$   OR   $\sqrt[3]{{\frac{{3.81971}}{{10.2832 \ldots }}}}$     (A1)

$r = 0.719{\text{ (m)}}$     (AG)

Note: The rounded and unrounded or formulaic answers must be seen for the final (A1) to be awarded. The use of $3.14$ gives an unrounded answer of $r = 0.719039 \ldots$.

[2 marks]

$0.75 = \frac{{\pi  \times {{(0.719)}^2}l}}{2}$     (M1)

Note: Award (M1) for substituting $0.719$ into their volume formula. Follow through from part (b).

$l = 0.924{\text{ (m)}}$  $(0.923599 \ldots )$     (A1)(ft)(G2)

[2 marks]

$T = (2\pi  + 4) \times 0.719 + \frac{6}{{\pi {{(0.719)}^2}}}$     (M1)

Notes: Award (M1) for substituting $0.719$ in their expression for $T$. Accept alternative methods, for example substitution of their $l$ and $0.719$ into their part (a) (for which the answer is $11.08961024$). Follow through from their answer to part (a).

$= 11.1{\text{ (m)}}$   $(11.0880 \ldots )$     (A1)(ft)(G2)