Question

Find the greatest value of $xyz$ for positive value of $x,y,z$ subject to the condition $xy+yz+xz=12$.

Answer 1

In this question we have been given an expression and we have to find the value of $xyz$ which is the multiplication of the three terms $x,y$ and $z$, given the condition $xy+yz+xz=12$. We will use the property of the mean that arithmetic mean is greater than the geometric mean. We will use the formula of the arithmetic mean which is given by $A.M=\dfrac{Sum\text{ }of\text{ }terms}{n}$ and also use the formula for the geometric mean which is given by $G.M=\sqrt[n]{product\text{ }of\text{ }terms}$, in both the cases $n$ represents the number of terms. we will create an inequality and get the required solution.

Complete step by step answer:
We have condition given to us as:
$\Rightarrow xy+yz+xz=12\to \left( 1 \right)$
We can see we have the three terms as $xy,yz$ and $zx$.
On writing the arithmetic mean of the terms, we get:
$\Rightarrow \dfrac{xy+yz+zx}{3}$
On writing the geometric mean of the terms, we get:
$\Rightarrow \sqrt[3]{xy\times yz\times xz}$
Now we know the property that $A.M\ge G.M$ therefore, we get:
$\Rightarrow \dfrac{xy+yz+zx}{3}\ge \sqrt[3]{xy\times yz\times xz}$
Now from equation $\left( 1 \right)$, we can substitute and write:
$\Rightarrow \dfrac{12}{3}\ge \sqrt[3]{xy\times yz\times xz}$
On simplifying, we get:
$\Rightarrow 4\ge \sqrt[3]{xy\times yz\times xz}$
On multiplying the terms in the right-hand side, we get:
$\Rightarrow 4\ge \sqrt[3]{{{\left( xyz \right)}^{2}}}$
On cubing both the sides, we get:
$\Rightarrow 64\ge {{\left( xyz \right)}^{2}}$
On taking the square root on both the sides, we get:
$\Rightarrow \sqrt{64}\ge xyz$
Now we know that $\sqrt{64}=8$ therefore, on substituting, we get:
$\Rightarrow 8\ge xyz$
Therefore, we get the inequality as $xyz$ is always less than or equal to $8$ which implies that the maximum value of $xyz=8$, which is the required solution.

Note: In this question we used the property that $A.M\ge G.M$. The actual property is $A.M\ge G.M\ge H.M$, where $H.M$ stands for the harmonic mean which has a formula as $H.M=\dfrac{n}{\dfrac{1}{{{x}_{1}}}+\dfrac{1}{{{x}_{2}}}+\dfrac{1}{{{x}_{3}}}....\dfrac{1}{{{x}_{n}}}}$, where ${{x}_{1}},{{x}_{2}},{{x}_{3}}....{{x}_{n}}$ represent the individual terms in the distribution and $n$ is the total number of terms.