A van is carrying a large amount of money in cash to deposit it in two ATM machines on a hill…

When the van starts moving, $x = 0$. $h(0) = 2(0)^3 - 18(0)^2 + 48(0) + 3 = 3$ (in 100 m). Height $= 3 \times 100\text{ m} = 300\text{ m}$. Hence, the van is at a height of 300 m when it starts moving.

The turning points occur where $h'(x) = 0$. $h'(x) = 6x^2 - 36x + 48 = 6(x^2 - 6x + 8) = 6(x - 2)(x - 4)$. Setting $h'(x) = 0$ gives $x = 2$ and $x = 4$. Therefore, the two ATM machines are located at $x = 2$ and $x = 4$.

At $x = 2$, $h(2) = 2(2)^3 - 18(2)^2 + 48(2) + 3 = 16 - 72 + 96 + 3 = 43$ (in 100 m) $= 4300\text{ m}$. At $x = 4$, $h(4) = 2(4)^3 - 18(4)^2 + 48(4) + 3 = 128 - 288 + 192 + 3 = 35$ (in 100 m) $= 3500\text{ m}$. Difference between heights $= 4300\text{ m} - 3500\text{ m} = 800\text{ m}$ (or $8$ units of $100\text{ m} = 0\cdot 8\text{ km}$).

From subpart (iii), the difference between the heights of the two ATM machines is $800\text{ m} = 0\cdot 8\text{ km}$. Since $0\cdot 8\text{ km} \le 1\text{ km}$ (the difference is not greater than 1 km), an extra armed security guard will not be required.

Evaluating $h(x)$ at the critical points and endpoints in $[0, 4]$: $h(0) = 3$ $h(2) = 43$ $h(4) = 35$ Absolute maximum value $= 43$ (at $x = 2$). Absolute minimum value $= 3$ (at $x = 0$).