First slide

Means - A.M/G.M/H.M

Question

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

Moderate

A

2180
B

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

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

2285

Solution

Since, $\frac{\frac{2 p}{3} + \frac{2 p}{3} + \frac{2 p}{3} + \underset{5 times}{\underset{⏟}{\frac{3 q}{5} + \dots + \frac{3 q}{5}}} + \underset{7 times}{\underset{⏟}{\frac{4 r}{7} + \dots + \frac{4 r}{7}}}}{15}$
$\geq \sqrt[15]{{(\frac{2 p}{3})}^{3} {(\frac{3 q}{5})}^{5} {(\frac{4 r}{7})}^{7}}$ [AM $\geq$ GM]
$\Rightarrow p^{3} q^{5} r^{7} \frac{2^{3} 3^{5} 4^{7}}{3^{3} 5^{5} 7^{7}} \leq 1 \Rightarrow p^{3} q^{5} r^{7} \leq \frac{5^{5} 7^{7}}{2^{3} 3^{2} 4^{7}} \leq \frac{5^{5} 7^{7}}{2^{17} 9}$

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