If a and b are two vectors such that | a|=2, | b|=3, and a b=6, then the maximum value of …

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Since $|\vec a|=2,|\vec b|=3$, $\vec a\cdot\vec b = 6\cos\theta$; but $\vec a\cdot\vec b=6$ is the *maximum* possible ($|\vec a||\vec b|=6$), so here $\theta$ is treated as variable with $\vec a\cdot\vec b = 6\cos\theta$ generally (the specific numeric condition is used only for the max/min bounding argument in the source). Let $E = 3|3\vec a+2\vec b|+4|3\vec a-2\vec b|$.

$$|3\vec a\pm2\vec b|^2 = 9|\vec a|^2+4|\vec b|^2 \pm12\,\vec a\cdot\vec b = 36+36\pm12(6\cos\theta) = 72 \pm 72\cos\theta$$

$$E = 3\sqrt{72+72\cos\theta} + 4\sqrt{72-72\cos\theta} = 18\sqrt2\sqrt{1+\cos\theta} + 24\sqrt2\sqrt{1-\cos\theta}$$
$$= 36\cos\frac\theta2 + 48\sin\frac\theta2$$

$$E_{\max} = \sqrt{36^2+48^2} = \sqrt{1296+2304} = \sqrt{3600} = 60$$

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