Question

If,
a, b, c belongs to R+
a + b + c = 18

find max value of a²b³c⁴?​

Answer 1

Step-by-step explanation:

$\huge\bf{\mid{\overline{\underline{ANSWER:-}}\mid}}$

$\huge \bold{GIVEN:-} \\ \\ a,b,c \: \: \in \: \: R ^{ + } \\ \\ a + b + c = 18 \\ \\ \large \red{To \: find:-} \\ \\ {a}^{2} {b}^{3} {c}^{4} = ? \: (max)$

$a + b + c = 18 \\ \\ \frac{a}{2} + \frac{a}{2} + \frac{b}{3} + \frac{b}{3} + \frac{b}{ 3} + \frac{c}{4} + \frac{c}{4} + \frac{c}{4} + \frac{c}{4} = 18$

$A. M \geq G. M$

$\frac{\frac{a}{2} + \frac{a}{2} + \frac{b}{3} + \frac{b}{3} + \frac{b}{ 3} + \frac{c}{4} + \frac{c}{4} + \frac{c}{4} + \frac{c}{4} }{9} \geq \\ \\ \sqrt[9]{\frac{a}{2} \times \frac{a}{2} \times \frac{b}{3} \times \frac{b}{3} \times \frac{b}{ 3} \times \frac{c}{4} \times \frac{c}{4} \times \frac{c}{4} \times \frac{c}{4} } \\ \\ {2}^{9} \geq( \frac{ {a}^{2} }{4} \times \frac{ {b}^{3} }{27} \times \frac{ {c}^{4} }{ {4}^{4} } ) \\ \\ \huge \boxed{ {a}^{2} \times {b}^{3} \times {c}^{4} \leq {2}^{19} \times {3}^{2} }$

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