Find Maximum Value of Vector Expression with Given Magnitudes

To find the maximum value of the given expression, let us first compute the magnitudes $|3\vec{a} + 2\vec{b}|$ and $|3\vec{a} - 2\vec{b}|$.

Given: $|\vec{a}| = 2 \quad \text{and} \quad |\vec{b}| = 3$

We have: $|3\vec{a}|^2 = 9|\vec{a}|^2 = 9(2)^2 = 36$ $|2\vec{b}|^2 = 4|\vec{b}|^2 = 4(3)^2 = 36$

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$, where $\theta \in [0, \pi]$. The dot product of $3\vec{a}$ and $2\vec{b}$ is: $3\vec{a} \cdot 2\vec{b} = 6(\vec{a} \cdot \vec{b}) = 6|\vec{a}||\vec{b}|\cos\theta = 6(2)(3)\cos\theta = 36\cos\theta$

Now, using the properties of vector magnitude: $|3\vec{a} + 2\vec{b}|^2 = |3\vec{a}|^2 + |2\vec{b}|^2 + 2(3\vec{a} \cdot 2\vec{b})$ $|3\vec{a} + 2\vec{b}|^2 = 36 + 36 + 2(36\cos\theta) = 72(1 + \cos\theta)$

Using the half-angle trigonometric identity $1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right)$: $|3\vec{a} + 2\vec{b}|^2 = 72 \cdot 2\cos^2\left(\frac{\theta}{2}\right) = 144\cos^2\left(\frac{\theta}{2}\right)$

Similarly, for $|3\vec{a} - 2\vec{b}|$: $|3\vec{a} - 2\vec{b}|^2 = |3\vec{a}|^2 + |2\vec{b}|^2 - 2(3\vec{a} \cdot 2\vec{b})$ $|3\vec{a} - 2\vec{b}|^2 = 36 + 36 - 2(36\cos\theta) = 72(1 - \cos\theta)$

Using the identity $1 - \cos\theta = 2\sin^2\left(\frac{\theta}{2}\right)$: $|3\vec{a} - 2\vec{b}|^2 = 72 \cdot 2\sin^2\left(\frac{\theta}{2}\right) = 144\sin^2\left(\frac{\theta}{2}\right)$

Since $\theta \in [0, \pi]$, we have $\frac{\theta}{2} \in \left[0, \frac{\pi}{2}\right]$, where both $\sin\left(\frac{\theta}{2}\right) \ge 0$ and $\cos\left(\frac{\theta}{2}\right) \ge 0$. Taking the square root on both sides: $|3\vec{a} + 2\vec{b}| = 12\cos\left(\frac{\theta}{2}\right)$ $|3\vec{a} - 2\vec{b}| = 12\sin\left(\frac{\theta}{2}\right)$

Now, let $E(\theta)$ be the expression whose maximum value we wish to find: $E(\theta) = 3|3\vec{a} + 2\vec{b}| + 4|3\vec{a} - 2\vec{b}|$ $E(\theta) = 3 \cdot 12\cos\left(\frac{\theta}{2}\right) + 4 \cdot 12\sin\left(\frac{\theta}{2}\right)$ $E(\theta) = 36\cos\left(\frac{\theta}{2}\right) + 48\sin\left(\frac{\theta}{2}\right)$

The maximum value of an expression of the form $A\cos x + B\sin x$ is given by $\sqrt{A^2 + B^2}$. Here, $A = 36$ and $B = 48$: $E_{\text{max}} = \sqrt{36^2 + 48^2} = \sqrt{12^2(3^2 + 4^2)} = 12 \sqrt{9 + 16} = 12 \times 5 = 60$

This maximum is achieved when $\tan\left(\frac{\theta}{2}\right) = \frac{48}{36} = \frac{4}{3}$, which yields a valid angle $\frac{\theta}{2} \in \left[0, \frac{\pi}{2}\right]$.

Hence, the maximum value is 60.

Correct Option: C