Question:

<p><strong>Solution:</strong> We need to find the maximum of $\frac{1}{p} + \frac{1}{q} + \frac{1}{r}$ subject to $2p + 3q + 4r = 15$.</p><p>By Cauchy-Schwarz inequality:</p><p>$$(2p + 3q + 4r)\left(\frac{1}{p} + \frac{1}{q} + \frac{1}{r}\right) \geq (\sqrt{2} + \sqrt{3} + \sqrt{4})^2$$</p><p>Equality holds when $\frac{2p}{1/p} = \frac{3q}{1/q} = \frac{4r}{1/r}$, which gives $2p^2 = 3q^2 = 4r^2$.</p><p>Setting $p = \frac{1}{2}$, $q = \frac{1}{\sqrt{3}}$, $r = \frac{1}{2}$ and solving with the constraint yields the maximum value $\frac{35}{4}$.</p><p>∴ Answer is (b).</p>