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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
VerifiedVerified
454.5k+ views
Hint: 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 \[a,b,c\] .