Question

Let $a+b+c=5\left( a,b,c >0 \right)$ and ${{x}^{2}}{{y}^{3}}=243\left( x >0,y >0 \right)$ ,
The greatest value of $a{{b}^{3}}c$ is –

& \left( a \right)3 \\\ & \left( b \right)9 \\\ & \left( c \right)27 \\\ & \left( d \right)81 \\\ \end{aligned}$$

Answer 1

Problems like these are a bit difficult to solve and understand, but can be done easily once we get all the underlying concepts behind the problem. This particular problem requires some previous knowledge of Arithmetic mean, Geometric mean and Harmonic mean and their various relations. Using all these we can pretty easily solve this problem. We must keep in mind one fact that, Arithmetic mean is always greater than or equal to Geometric mean, and Geometric mean is always greater than or equal to Harmonic mean. In the mathematical form we write, $AM\ge GM\ge HM$ .

Complete step by step answer:
Now we start off with the solution to the given problem by splitting the part $b$ into three equal parts because we are needed to find the value of $a{{b}^{3}}c$ and it contains a part ${{b}^{3}}$ . The equal parts of $b$ are thus $\dfrac{b}{3},\dfrac{b}{3},\dfrac{b}{3}$ . Now, here we apply the concept that Arithmetic mean is greater than or equal to geometric mean. We write it down as,
$\dfrac{a+\dfrac{b}{3}+\dfrac{b}{3}+\dfrac{b}{3}+c}{5}\ge \sqrt[5]{a\times \dfrac{b}{3}\times \dfrac{b}{3}\times \dfrac{b}{3}\times c}$
Here the left hand side is the Arithmetic mean and the right hand side is the Geometric mean. We now evaluate it as,
$=\dfrac{a+b+c}{5}\ge \sqrt[5]{\dfrac{a\times {{b}^{3}}\times c}{27}}$
Now we have $a+b+c=5$ , putting this in our intermediate equation we get,
$=1\ge \sqrt[5]{\dfrac{a\times {{b}^{3}}\times c}{27}}$
Now doing the fifth power on both the sides of the equation we get,

& =1\ge \dfrac{a\times {{b}^{3}}\times c}{27} \\\ & =\dfrac{a\times {{b}^{3}}\times c}{27}\le 1 \\\ & =a{{b}^{3}}c\le 27 \\\ \end{aligned}$$ So we get the maximum value of $$a{{b}^{3}}c$$ as $$27$$ . So our answer to the problem is option $$\left( c \right)$$ . **Note:** Such types of problems require some fair knowledge of chapters like Arithmetic progression, Geometric progression and Harmonic progression. Here we need to split the particular variable cautiously because any mistake made while splitting will affect our result and lead to a wrong solution. We also need to keep in mind all the possible relations of Arithmetic mean and Geometric mean or else we would not be able to proceed with the problem.