Question

Find the greatest value of ${a^2}{b^3}{c^2}$, if $a + b + c = 3$ and $a > 0,b > 0,c > 0$. Choose the correct answer.
(A) ${3^2}{2^3}{2^2}$
(B) ${3^2}{2^3}{7^2}$
(C) $\dfrac{{{3^{10}} \times {2^4}}}{{{7^7}}}$
(D) $\dfrac{{{3^4} \times {2^{10}}}}{{{7^4}}}$

Answer 1

In this question we will use the concept of arithmetic mean and geometric mean to find the greatest value of the given term. In this we consider that arithmetic mean is greater than or equal to geometric mean. We write the term given in the form of arithmetic mean and geometric mean. Now we find the greatest value of the given term.

Complete step-by-step answer:
As it is given in the question $a + b + c = 3$.
In this we will use the arithmetic mean and geometric mean to find the greatest value of ${a^2}{b^3}{c^2}$.
We consider that $arithmetic{\text{ }}mean{\text{ }} \geqslant {\text{ }}geometric{\text{ }}mean$.
Now we write the given equation $a + b + c = 3$in the form of arithmetic mean, we get
$\Rightarrow \dfrac{a}{2} + \dfrac{a}{2} + \dfrac{b}{3} + \dfrac{b}{3} + \dfrac{b}{3} + \dfrac{c}{2} + \dfrac{c}{2} = 3$
Now we also write the given equation $a + b + c = 3$ in the form of geometric mean, we get
$\Rightarrow {\left( {\dfrac{{{a^2}}}{{{2^2}}} \times \dfrac{{{b^3}}}{{{3^3}}} \times \dfrac{{{c^2}}}{{{2^2}}}} \right)^{\dfrac{1}{7}}}$
Now we will use the relation between arithmetic mean and geometric mean as
$A.M \geqslant G.M$
Formula of $A.M = \dfrac{{sum{\text{ }}of{\text{ }}the{\text{ }}numbers}}{{number{\text{ }}of{\text{ }}terms}}$
Now we put the values in the formula we get,
$A.M = \dfrac{{\dfrac{a}{2} + \dfrac{a}{2} + \dfrac{b}{3} + \dfrac{b}{3} + \dfrac{b}{3} + \dfrac{c}{2} + \dfrac{c}{2}}}{7}$
We put the values of the equation in the above relation we get,
$\Rightarrow \dfrac{{\dfrac{a}{2} + \dfrac{a}{2} + \dfrac{b}{3} + \dfrac{b}{3} + \dfrac{b}{3} + \dfrac{c}{2} + \dfrac{c}{2}}}{7} \geqslant {\left( {\dfrac{{{a^2}}}{{{2^2}}} \times \dfrac{{{b^3}}}{{{3^3}}} \times \dfrac{{{c^2}}}{{{2^2}}}} \right)^{\dfrac{1}{7}}}$
Now we put the value of $\dfrac{a}{2} + \dfrac{a}{2} + \dfrac{b}{3} + \dfrac{b}{3} + \dfrac{b}{3} + \dfrac{c}{2} + \dfrac{c}{2} = 3$ in the equation, we get,
$\Rightarrow \dfrac{3}{7} \geqslant {\left( {\dfrac{{{a^2}}}{{{2^2}}} \times \dfrac{{{b^3}}}{{{3^3}}} \times \dfrac{{{c^2}}}{{{2^2}}}} \right)^{\dfrac{1}{7}}}$
Now we multiply the equation by the power 7, we get
$\Rightarrow {\left( {\dfrac{3}{7}} \right)^7} \geqslant {\left( {\dfrac{{{a^2}}}{{{2^2}}} \times \dfrac{{{b^3}}}{{{3^3}}} \times \dfrac{{{c^2}}}{{{2^2}}}} \right)^{\dfrac{1}{7} \times 7}}$
By solving the above equation we get,
$\Rightarrow \dfrac{{{3^7}}}{{{7^7}}} \geqslant \dfrac{{{a^2}}}{{{2^2}}} \times \dfrac{{{b^3}}}{{{3^3}}} \times \dfrac{{{c^2}}}{{{2^2}}}$
Now in the denominator of R.H.S we apply the formula as
$\Rightarrow {a^m} \times {a^n} = {a^{m + n}}$, where $a = 2,m = 2,n = 2$ , we get
$\Rightarrow \dfrac{{{3^7}}}{{{7^7}}} \geqslant \dfrac{{{a^2} \times {b^3} \times {c^2}}}{{{2^4} \times {3^3}}}$
Now we move the constant value in the denominator of R.H.S to L.H.S, we get

$\Rightarrow \dfrac{{{3^7} \times {2^4} \times {3^3}}}{{{7^7}}} \geqslant {a^2}{b^3}{c^2}$
Now in the numerator of L.H.S we apply the formula as
$\Rightarrow {a^m} \times {a^n} = {a^{m + n}}$, where $a = 3,m = 7,n = 3$ , we get
$\Rightarrow \dfrac{{{3^{7 + 3}} \times {2^4}}}{{{7^7}}} \geqslant {a^2}{b^3}{c^2}$
$\Rightarrow \dfrac{{{3^{10}} \times {2^4}}}{{{7^7}}} \geqslant {a^2}{b^3}{c^2}$

So, the greatest value of ${a^2}{b^3}{c^2}$ is $\dfrac{{{3^{10}} \times {2^4}}}{{{7^7}}}$.

So, the correct answer is “Option C”.

Note: In these type of questions we should write the given equation in the form of arithmetic mean and geometric mean. And then we should compare both arithmetic mean and geometric mean to find the greatest value of the given term. We also remember to calculate the total number of terms in the given equation after converting it in the form of arithmetic mean and geometric mean.