If $2 p + 3 q + 4 r = 15$ then the maximum value of $p^{3} q^{5} r^{7}$ is?

2180

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\frac{5^{4} \cdot 3^{5}}{2^{15}}

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\frac{5^{5} \cdot 7^{7}}{2^{17} \cdot 9}

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2285

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Solution

The correct option is C $\frac{5^{5} \cdot 7^{7}}{2^{17} \cdot 9}$
This can be done very easily using the AM - GM Inequality

$\frac{\frac{3 \times 2 p}{3} + \frac{5 \times 3 q}{5} + \frac{7 \times 4 r}{7}}{15} \geq \sqrt{(\frac{2 p}{3})^{3} (\frac{3 q}{5})^{5} (\frac{4 r}{7})^{7}}$
$\to p^{3} q^{5} r^{7} \leq \frac{5^{5} 7^{7}}{2^{17} .9}$

So C is the correct option

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