A tank obliquely cut

When a plane cuts two parallel planes, the intersection is a pair of parallel lines, hence ABCD is a parallelogram, thus M is the midpoint of AC and BD. EFGH is a rectangle, thus N is the midpoint of EG and HF. Consequently, MN is the midsegment of both trapeziums ACGE and BDHF.

Since the length of the midsegment of a trapezium is half the sum of the lengths of the two parallel sides we get: $\left| {\overline {MN} } \right| = \frac{{\left| {\overline {DH} } \right| + \left| {\overline {BF} } \right|}}{2} \Rightarrow \left| {\overline {MN} } \right| = \frac{{4 + 6}}{2} \Rightarrow \left| {\overline {MN} } \right| = 5$ $\left| {\overline {MN} } \right| = \frac{{\left| {\overline {AE} } \right| + \left| {\overline {CG} } \right|}}{2} \Rightarrow 5 = \frac{{\left| {\overline {AE} } \right| + 7}}{2} \Rightarrow \left| {\overline {AE} } \right| = 3$ The tank can hold liquid up to the level of point A, hence the maximum volume is $2 \times 5 \times 3 = \boxed{30}$ . $\\~\\~\\$ Clarification : obviously, the term “ trapezium ” is used here in the way Proclus and Archimedes defined it, i.e. the quadrilateral with a pair of parallel sides.

Since the top plane $ABCD$ cuts two parallel planes, $AD || BC$ , which means $AD$ and $BC$ have the same slope. Therefore, $\frac{7 - 6}{5} = \frac{4 - x}{5}$ , which solves to $x = 3$ .

The maximum volume of liquid the tank can hold is then $V = 5 \cdot 2 \cdot 3 = \boxed{30}$ .