4. Given that point $P(x, y)$ lies on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$, find the maximum value of $2 x-y$.

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4. Let $\vec{a}=\left(\frac{x}{2}, \frac{y}{3}\right), \vec{b}=(4,-3)$, then
$$
1=\frac{x^{2}}{4}+\frac{y^{2}}{9}=|\vec{a}|^{2} \geqslant \frac{(\vec{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{(2 x-y)^{2}}{4^{2}+(-3)^{2}}=\frac{(2 x-y)^{2}}{25} \text {. }
$$
$(2 x-y)^{2} \leqslant 25, 2x-y$'s maximum value is 5.