The resistance of the heating element is:
\[R = \rho \frac{\ell}{A}.\]
The power is inversely proportional to the length:
\[P \propto \frac{1}{\ell}.\]
From the relation \(P_1 \times t_1 = P_2 \times t_2\):
\[\frac{P_1}{P_2} = \frac{t_2}{t_1} = \frac{15}{20}.\]
Substituting \(P \propto \frac{1}{\ell}\):
\[\frac{\ell_2}{\ell_1} = \frac{t_2}{t_1} = \frac{15}{20} = \frac{3}{4}.\]
Thus, the new length should be decreased to:
\[\ell_2 = \frac{3}{4} \ell_1.\]
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: