Question

If $x,y,z\in R^{+}$ such that $x^3{y}^2{z}^4=7$, then the least value of $2x+5y+3z$ is

• A. ${\left(\frac{525}{128}\right)}^{\frac{1}{9}}$
• B. $3{\left(\frac{525}{128}\right)}^{\frac{1}{9}}$
• C. $9{\left(\frac{525}{128}\right)}^{\frac{1}{9}}$
• D. None of the above.

Answer 1

The correct option is C $9{\left(\frac{525}{128}\right)}^{\frac{1}{9}}$

given $x^3{y}^2{z}^4=7$

we know that $A.M.\ge G.M.$

$⟹\frac{\frac{2x}{3}+\frac{2x}{3}+\frac{2x}{3}+\frac{5y}{2}+\frac{5y}{2}+\frac{3z}{4}+\frac{3z}{4}+\frac{3z}{4}+\frac{3z}{4}}{9}\ge \sqrt[9]{92^3\ast \frac{x}{3}\ast \frac{x}{3}\ast \frac{x}{3}\ast 5^2\ast \frac{y}{2}\ast \frac{y}{2}\ast 3^4\ast \frac{z}{4}\ast \frac{z}{4}\ast \frac{z}{4}\ast \frac{z}{4}}$

$⟹\frac{2\left(\frac{3x}{3}\right)+5\left(\frac{2y}{2}\right)+3\left(\frac{4z}{4}\right)}{9}\ge \sqrt[9]{92^3\ast 5^2\ast 3^4\ast \frac{x^3}{3^3}\ast \frac{y^2}{2^2}\ast \frac{z^4}{2^8}}$

$⟹\frac{2x+5y+3z}{9}\ge \sqrt[9]{9\frac{2^3\ast 5^2\ast 3^4\ast x^3{y}^2{z}^4}{2^{10}\ast 3^3}}$

$⟹\frac{2x+5y+3z}{9}\ge \sqrt[9]{9\frac{5^2\ast 3\ast 7}{2^7}}$

$⟹2x+5y+3z\ge 9\sqrt[9]{9\frac{525}{128}}$

$⟹2x+5y+3z\ge 9(\frac{525}{128}{)}^{\frac{1}{9}}$

Therefore the least value of $2x+5y+3z$ is $9(\frac{525}{128}{)}^{\frac{1}{9}}$