Question

The greatest value of $xyz$ for +ve values of $x,y,z$; where $yz + zx + xy = 12$ is
A.$4$
B.$6$
C.$8$
D.$10$

Answer 1

Here we are asked to find the greatest value of$xyz$, for positive values of $x,y,z$ . And it is also given that$yz + zx + xy = 12$. We will solve this problem by using arithmetic mean and geometric mean. We know that arithmetic mean $\geqslant$the geometric mean. Using this relation by taking $yz,zx\& xy$ as a separate term to find A.M and G.M then we will find the greatest value of$xyz$.
Formula: The formulas that we need to know to solve this problem:
Arithmetic mean: $\dfrac{{a + b + c}}{3}$
Geometric mean: $\sqrt[3]{{abc}}$
$A.M. \geqslant G.M.$

Complete step by step answer:
It is given that$yz + zx + xy = 12$. We aim to find the greatest value of the term$xyz$, for positive values of$x,y,z$.
We will solve this problem by using arithmetic mean and geometric mean. Let us take $a = yz,b = zx$and$c = xy$.
Now we will find the arithmetic mean and the geometric mean of $a,b,c$
Arithmetic mean:
We know that the arithmetic means of a given number of observations will be equal to the sum of all the observations divided by the total number of observations.
Here $a = yz,b = zx$ and$c = xy$ then the arithmetic means of $a,b,c$ is $\dfrac{{a + b + c}}{3}$.
Now let us substitute the values of$a,b,c$.
\dfrac{{a + b + c}}{3}$$$$ = \dfrac{{yz + zx + xy}}{3}
From the given data we have that $yz + zx + xy = 12$let us substitute it in the above
$= \dfrac{{12}}{3}$
\dfrac{{a + b + c}}{3}$$$$ = 4
Thus, we have found that the arithmetic means of $a,b,c$ is $4$.
Geometric mean:
We know that the geometric mean of a given number of data is equal to the cubic root of the product of all data.
Here $a = yz,b = zx$and $c = xy$ then the geometric mean of $a,b,c$ is $\sqrt[3]{{abc}}$.
Now let us substitute the values of $a,b,c$.
\sqrt[3]{{abc}}$$$$ = \sqrt[3]{{(yz)(zx)(xy)}}
$= \sqrt[3]{{{x^2}{y^2}{z^2}}}$
\sqrt[3]{{abc}}$$$$ \Rightarrow \sqrt[3]{{{{(xyz)}^2}}} \leqslant 4
Thus, we have found that the geometric mean of $a,b,c$is $\sqrt[3]{{{{(xyz)}^2}}}$.
The relation between the arithmetic mean and geometric mean is $A.M. \geqslant G.M.$.
Let us substitute the values of A.M. and G.M. in this relation.
$4 \geqslant \sqrt[3]{{{{(xyz)}^2}}}$
Let us re-write the above relation for our convenience.
$\Rightarrow \sqrt[3]{{{{(xyz)}^2}}} \leqslant 4$
Now let’s raise the power to three on both sides.
$\Rightarrow {\left( {\sqrt[3]{{{{(xyz)}^2}}}} \right)^3} \leqslant {4^3}$
$\Rightarrow {(xyz)^2} \leqslant 64$
Now let us take square root on both sides of the above inequality.
$\Rightarrow \sqrt {{{(xyz)}^2}} \leqslant \sqrt {64}$
$\Rightarrow xyz \leqslant 8$
Thus, the greatest value of $xyz$ is $8$ .
Now let us see the options, option (a) $4$ is an incorrect answer as we got that $8$ as the greatest value of $xyz$
Option (b) $6$ is an incorrect answer as we got that $8$ as the greatest value of $xyz$
Option (c) $8$ is the correct option as we got the same answer in our calculation above
Option (d) $10$ is an incorrect answer as we got that $8$ as the greatest value of $xyz$

Note:
Here the arithmetic mean is denoted as A.M. and the geometric mean is denoted as G.M. Since the terms $x,y,z$ are all positive numbers then their product will be a numeric value thus we have considered their products as separate numerals $a,b,c$ .