The greatest value of $a b^{3} c$ is -

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 27
Given, $a + b + c = 5 (a, b, c > 0)$
Thus using $A M \geq G M$
$= \frac{a + \frac{b}{3} + \frac{b}{3} + \frac{b}{3} + c}{5} \geq {(\frac{a b^{3} c}{27})}^{1 / 5}$
$\Rightarrow 1^{5} \geq \frac{a b^{3} c}{27} \Rightarrow a b^{3} c \leq 27$

Suggest Corrections

Similar questions

If $A x + B y = 1$ is a normal to the curve $a y = x^{2}$ , then

P (A) = 0.6

P (A \cap B) = 0.4

P (B)

\frac{k}{10}

k

sin x cos x

(sin α cos α)

{sin}^{4} q + {cos}^{4} q

Join BYJU'S Learning Program