If \( x \) is jointly proportional to the cube of \( y \) and the fourth power of \( z \), we can write:

\[ x = k \cdot y^3 \cdot z^4 \]

where \( k \) is the constant of proportionality.

Let's analyze the changes when \( y \) is halved and \( z \) is doubled:

1. Original Value:
\[ x = k \cdot y^3 \cdot z^4 \]

2. New Value:
\[ y \text{ is halved } \implies y \rightarrow \frac{y}{2} \]
\[ z \text{ is doubled } \implies z \rightarrow 2z \]

The new value of \( x \) becomes:
\[ x' = k \cdot \left(\frac{y}{2}\right)^3 \cdot (2z)^4 \]

Simplify this:
\[ x' = k \cdot \frac{y^3}{2^3} \cdot 2^4 \cdot z^4 \]
\[ x' = k \cdot \frac{y^3}{8} \cdot 16 \cdot z^4 \]
\[ x' = \frac{16}{8} \cdot k \cdot y^3 \cdot z^4 \]
\[ x' = 2 \cdot k \cdot y^3 \cdot z^4 \]
\[ x' = 2x \]

Thus, \( x \) is increased by a factor of 2.

So the ratio of the increase is 2:1.

The correct answer is:

B. 2:1 increase