Question

If \[a,b,c\in {{R}^{+}}\] such that \[ab{{c}^{2}}\] has the greatest value A.\[\dfrac{1}{64}\] , then
B.\[a=b=\dfrac{1}{2},c=\dfrac{1}{4}\]
C.\[a=b=\dfrac{1}{4},c=\dfrac{1}{2}\]
D.\[a=b=c=\dfrac{1}{3}\]
None of the above

Answer 1

Hint: We know the formula that, \[A.M\ge G.M\] if \[a,b,c\in {{R}^{+}}\] . Use this formula for \[a,b,\dfrac{c}{2},\dfrac{c}{2}\] . We know that the G.M has the greatest value when equality holds and equality only holds when all the positive numbers are equal, that is \[a=b=\dfrac{c}{2}\] . Now, we have \[A.M=G.M\] and \[a=b=\dfrac{c}{2}\] . Now, solve it further.

Complete step-by-step answer:
According to the question, we have \[a,b,c\in {{R}^{+}}\] and greatest value of \[ab{{c}^{2}}\] is \[\dfrac{1}{64}\] .
It means that a, b, and c are the real positive numbers.
We know the formula that if a, b, and c are the real positive numbers then, \[A.M\ge G.M\] .
If it holds the equality then, G.M will have the greatest value and equality will hold only when all the numbers are equal.
Let us take \[a,b,\dfrac{c}{2},\dfrac{c}{2}\] . As \[a,b,c\in {{R}^{+}}\] so \[a,b,\dfrac{c}{2},\dfrac{c}{2}\in {{R}^{+}}\]and now, use the formula \[A.M\ge G.M\] for
\[a,b,\dfrac{c}{2},\dfrac{c}{2}\in {{R}^{+}}\] .
\[\dfrac{a+b+\dfrac{c}{2}+\dfrac{c}{2}}{4}\ge {{\left( a.b.\dfrac{c}{2}.\dfrac{c}{2} \right)}^{\dfrac{1}{4}}}\]
On solving, we get
\[\dfrac{a+b+c}{4}\ge {{\left( \dfrac{ab{{c}^{2}}}{4} \right)}^{\dfrac{1}{4}}}\] ……………………………(1)
Greatest value of \[ab{{c}^{2}}\] is \[\dfrac{1}{64}\] .
For equation (1), equality will hold if and only if \[a=b=\dfrac{c}{2}\] and \[A.M=G.M\],
\[a=b=\dfrac{c}{2}\] ………………..(2)
\[A.M=G.M\] ………………(3)
From equation (2) and equation (3), we get
\[\begin{align}
  & \dfrac{a+b+c}{4}={{\left( \dfrac{1}{64\times 4} \right)}^{\dfrac{1}{4}}} \\
 & \Rightarrow \dfrac{a+b+c}{4}=\dfrac{1}{4} \\
 & \Rightarrow \dfrac{a+a+2a}{4}=\dfrac{1}{4} \\
 & \Rightarrow \dfrac{4a}{4}=\dfrac{1}{4} \\
 & \Rightarrow a=\dfrac{1}{4} \\
\end{align}\]
Using equation (2), we get
\[a=b=\dfrac{c}{2}=\dfrac{1}{4}\]
On solving,
\[\begin{align}
  & \dfrac{c}{2}=\dfrac{1}{4} \\
 & \Rightarrow c=\dfrac{1}{2} \\
\end{align}\]
So, \[a=b=\dfrac{1}{4}\] and \[c=\dfrac{1}{2}\] .
Hence, option (B) is the correct one.

Note:In this question, one may think to apply the formula \[A.M\ge G.M\] for \[a,b,\dfrac{c}{3},\dfrac{c}{3},\dfrac{c}{3}\] .
If we do so then we will get,
\[\begin{align}
  & \dfrac{a+b+\dfrac{c}{3}+\dfrac{c}{3}+\dfrac{c}{3}}{5}\ge {{\left( a.b.\dfrac{c}{3}.\dfrac{c}{3}.\dfrac{c}{3} \right)}^{\dfrac{1}{5}}} \\
 & \Rightarrow \dfrac{a+b+c}{5}\ge {{\left( \dfrac{ab{{c}^{3}}}{27} \right)}^{\dfrac{1}{5}}} \\
\end{align}\]
In the question, we don’t have any information for \[ab{{c}^{3}}\] . So, we can’t approach this question by this method. In order to use the information that is provided for \[ab{{c}^{2}}\] , we have to apply the formula \[A.M\ge G.M\] for \[a,b,\dfrac{c}{2},\dfrac{c}{2}\] .