Question

If $3x+7y+4z=21$, where $x,y,z$ are positive real numbers, then maximum value of $x^4{y}^5{z}^3$ is equal to

• A. $\frac{7^7\times 5^5\times 4^{-10}}{12}$
• B. $\frac{7^7\times 5^5\times 4^{10}}{12}$
• C. $\frac{7^6\times 5^7}{4^{11}\times 3}$
• D. $\frac{7^5\times 5^6}{4^{10}\times 3}$

Answer 1

The correct option is A $\frac{7^7\times 5^5\times 4^{-10}}{12}$

Consider the values $\left(\frac{3x}{4}\right)$ $4$ times $\left(\frac{7y}{5}\right)$$5$ times and $\left(\frac{4z}{3}\right)$ $3$ times

Using $AM.GM$ Inequality we, get

$\frac{\left(\frac{3x}{4}\right)\times 4+\left(\frac{7y}{5}\right)\times 5+\left(\frac{4z}{3}\right)\times 3}{4+5+3}\ge \sqrt[12]{12{\left(\frac{3x}{4}\right)}^4{\left(\frac{7y}{5}\right)}^5{\left(\frac{4z}{3}\right)}^3}$

$\frac{3x+7y+4z}{12}\ge \sqrt[12]{12{\left(\frac{3x}{4}\right)}^4{\left(\frac{7y}{5}\right)}^5{\left(\frac{4z}{3}\right)}^3.}$

${\left(\frac{21}{12}\right)}^{12}\ge {\left(\frac{3x}{4}\right)}^4{\left(\frac{7y}{5}\right)}^5{\left(\frac{4z}{3}\right)}^3$ [Raising to the ${12}^{th}$ power]

$x^4.y^5.z^3\le \frac{{21}^{12}.4^4.5^5.3^3}{{12}^{12}.3^4.7^5.4^3}$

$x^4.y^5.z^3\le \frac{7^{12}.3^{12}.3^3.2^8.5^5}{3^{12}.2^{24}.2^6.3^4.7^5}$

$x^4.y^5.z^3\le \frac{7^7.5^5}{2^{22}.3}=\frac{7^7.5^5}{4^{11}.3}$

$x^4.y^5.z^3\le \frac{7^7{.5}^5{.4}^{-10}}{12}$

Maximum value of $x^4.y^5.z^3=\frac{7^7{.5}^5{.4}^{-10}}{12}$