Figure 1 shows an open cylindrical tank. Figure 2 shows the cross-sectional view of the tank. The external radius of the tank is $r$ cm, the external height is $h$ cm and the tank is made with a material of thickness $a$ cm throughout. The internal volume of the tank is fixed at $1000\pi$ $\text{cm}^3$.

1. Determine the internal dimensions and use the fixed internal volume to express the external height $h$ in terms of $r$ and $a$.

The internal radius is $r - a$ and the internal height is $h - a$.
Given the internal volume is $1000\pi$:
$\pi(r - a)^2(h - a) = 1000\pi$
$h - a = \frac{1000}{(r - a)^2}$
$h = \frac{1000}{(r - a)^2} + a$

2. Express the total volume of the material $V$ as the difference between the external volume and the internal volume.

The volume of the material is the total external volume of the cylinder minus the empty internal volume.
$V = V_{\text{external}} - V_{\text{internal}}$
$V = \pi r^2 h - 1000\pi$

3. Substitute the expression for $h$ into $V$ and simplify to reach the given form, identifying $k$.

Substitute $h = \frac{1000}{(r - a)^2} + a$ into the equation for $V$:
$V = \pi r^2 \left[ \frac{1000}{(r - a)^2} + a \right] - 1000\pi$
$V = \frac{1000\pi r^2}{(r - a)^2} + \pi r^2 a - 1000\pi$
Group the terms with $1000\pi$:
$V = 1000\pi \left[ \frac{r^2}{(r - a)^2} - 1 \right] + \pi r^2 a$
Comparing this with $V = k\pi \left[ \frac{r^2}{(r-a)^2} - 1 \right] + \pi r^2 a$, we determine that $k = 1000$.

1. State the value of $k$ from part (a) and write the expression for $V$.

From the internal volume of $1000\pi\text{ cm}^3$ in part (a), the constant $k$ is $1000$.
$V = 1000\pi \left[ \frac{r^2}{(r-a)^2} - 1 \right] + \pi r^2 a$

2. Differentiate $V$ with respect to $r$.

Using the quotient rule on the first term:
$\frac{\mathrm{d}V}{\mathrm{d}r} = 1000\pi \left[ \frac{2r(r-a)^2 - r^2 \cdot 2(r-a)}{(r-a)^4} \right] + 2\pi a r$
$\frac{\mathrm{d}V}{\mathrm{d}r} = 1000\pi \left[ \frac{2r(r-a) - 2r^2}{(r-a)^3} \right] + 2\pi a r$
$\frac{\mathrm{d}V}{\mathrm{d}r} = 1000\pi \left[ \frac{2r^2 - 2ar - 2r^2}{(r-a)^3} \right] + 2\pi a r$
$\frac{\mathrm{d}V}{\mathrm{d}r} = -\frac{2000\pi a r}{(r-a)^3} + 2\pi a r$

3. Set the derivative to zero for minimum volume.

For $V$ to be a minimum, set $\frac{\mathrm{d}V}{\mathrm{d}r} = 0$:
$-\frac{2000\pi a r}{(r-a)^3} + 2\pi a r = 0$
Since $r > 0$ and $a > 0$, we can divide throughout by $2\pi a r$:
$1 - \frac{1000}{(r-a)^3} = 0$

4. Solve for $r$ in terms of $a$.

$\frac{1000}{(r-a)^3} = 1$
$(r-a)^3 = 1000$
Taking the cube root of both sides:
$r - a = 10$
$r = a + 10$