Question
If $\overrightarrow{a}$ and $\overrightarrow{b}$ are two vectors such that $|\overrightarrow{a}| = 2$ and $|\overrightarrow{b}| = 3$, then the maximum value of $3|(3\overrightarrow{a} + 2\overrightarrow{b})| + 4|(3\overrightarrow{a} - 2\overrightarrow{b})|$ is :
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Solution
Sol. [Key Concept: Write both magnitudes via $a \cdot \ b = 6cos\theta$; maximize $Acos(\theta/2) + Bsin(\theta/2) = \sqrt{A^{2} + B^{2}}$.]
$$E = 3\sqrt{9a^{2} + 4b^{2} + 12\overrightarrow{a} \cdot \overrightarrow{b}} + 4\sqrt{9a^{2} + 4b^{2} - 12\overrightarrow{a} \cdot \overrightarrow{b}} $$
$$= 3\sqrt{36 + 36 + 12 \times 6\cos\theta} + 4\sqrt{36 + 36 - 72\cos\theta} $$
$$= 3\sqrt{72 + 72\cos\theta} + 4\sqrt{72 - 72\cos\theta} $$
$$= 18\sqrt{2}\sqrt{1 + \cos\theta} + 24\sqrt{2}\sqrt{1 - \cos\theta} $$
$$= 18\sqrt{2}\sqrt{2}\cos\frac{\theta}{2} + 24\sqrt{2}\left( \sqrt{2}\sin\frac{\theta}{2} \right) $$
$$= 36\left( \cos\frac{\theta}{2} \right) + 48\sin\left( \frac{\theta}{2} \right) $$
$$E_{\text{max}\text{~}} = \sqrt{(36)^{2} + (48)^{2}} = \sqrt{6^{4} + 6^{2} \times 8^{2}} = 60 $$
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