The greatest value of $x^{2} y^{3} z^{4}$ , (if $x + y + z = 1, x, y, z > 0)$ is-

\frac{2^{9}}{3^{5}}

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\frac{2^{10}}{3^{15}}

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\frac{2^{15}}{3^{10}}

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\frac{2^{10}}{3^{10}}

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Solution

The correct option is B $\frac{2^{10}}{3^{15}}$
$x + y + z = 1 \Rightarrow 2. \frac{x}{2} + 3. \frac{y}{3} + 4. \frac{z}{4} = 1$

Using weighted mean $\frac{2. (\frac{x}{2}) + 3. (\frac{y}{3}) + 4. (\frac{z}{4})}{9} \geq ((\frac{x}{2})^{2} (\frac{y}{3})^{3} (\frac{z}{4})^{4})^{\frac{1}{9}}$

$(\frac{1}{9})^{9} \geq \frac{x^{2} y^{3} z^{4}}{2^{10} 3^{3}} \Rightarrow x^{2} y^{3} z^{4} \leq \frac{2^{10}}{3^{15}}$

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Similar questions

x + y + z = 1

x, y, z > 0

x^{2} y^{3} z^{4}

x = 111 \dots \dots 1 (20 digits),

y = 333 \dots \dots 3 (10 digits)

z = 222 \dots \dots 2 (10 digits)

x - z

2^{10}

3^{10}

f (x) = 2^{10} \cdot x + 1

g (x) = 3^{10} \cdot x - 1

(f o g) (x) = x

x + y + z = 1

x, y, z > 0

x^{2} y^{3} z^{4}

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