Let x, y, z be positive real numbers such tha | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $x + y + z = 12$ and $x^3y^4z^5 = (0.1)(600)^3$.Using the Weighted Arithmetic Mean-Geometric Mean Inequality:$\frac{3(\frac{x}{3}) + 4(\frac

Vedclass · Accepted Answer

$216$

Vedclass · Answer

(B) Given $x + y + z = 12$ and $x^3y^4z^5 = (0.1)(600)^3$.Using the Weighted Arithmetic Mean-Geometric Mean Inequality:$\frac{3(\frac{x}{3}) + 4(\frac{y}{4}) + 5(\frac{z}{5})}{3+4+5} \ge ((\frac{x}{3})^3 (\frac{y}{4})^4 (\frac{z}{5})^5)^{1/12}$$\frac{x+y+z}{12} \ge (\frac{x^3 y^4 z^5}{3^3 4^4 5^5})^{1/12}$$\frac{12}{12} \ge (\frac{x^3 y^4 z^5}{27 	imes 256 	imes 3125})^{1/12}$$1 \ge \frac{x^3 y^4 z^5}{21600000}$$x^3 y^4 z^5 \le 21600000 = 21.6 	imes 10^6 = (0.1)(600)^3$.Since the equality holds,we must have $\frac{x}{3} = \frac{y}{4} = \frac{z}{5} = k$.Then $x = 3k, y = 4k, z = 5k$.$3k + 4k + 5k = 12 \implies 12k = 12 \implies k = 1$.Thus,$x = 3, y = 4, z = 5$.Therefore,$x^3 + y^3 + z^3 = 3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216$.

Let $x, y, z$ be positive real numbers such that $x + y + z = 12$ and $x^3y^4z^5 = (0.1)(600)^3$. Then $x^3 + y^3 + z^3$ is equal to

Similar Questions

If $1, \log _9(3^{1-x}+2), \log _3(4 \cdot 3^x-1)$ are in $A.P.$,then $x$ equals

If the sum of the first $2n$ terms of $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of $57, 59, 61, \dots$,then $n$ is equal to

If for an arithmetic progression $S_{2n} = 2S_n$,then $S_{3n} / S_n = \dots$

If $n$ is a natural number,then $\left( \frac{n+1}{2} \right)^n \ge n!$ is true when

If the sum of the first $n$ terms of the series $\sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots$ is $435\sqrt{3}$,then $n$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module