Question

If \[x + y = 8\], then find the maximum value of \[{x^2}y\] .
A. \[\dfrac{{{\bf{2048}}}}{9}\]
B. \[\dfrac{{{\bf{2048}}}}{{81}}\]
C. \[\dfrac{{{\bf{2048}}}}{3}\]
D. \[\dfrac{{{\bf{2048}}}}{{27}}\]

Answer 1

Hint:We will divide the variables in the given equation with respect to their powers and later use the relation formula of arithmetic mean and Geometric mean.

Formula used:
Relation between the arithmetic mean and geometric mean \[AM \ge GM\]

Complete step by step Solution:
Here, we are given the equation
\[x + y = 8\]
Firstly, we will separate the variable \[x\] in 2 parts by dividing process
So, we get
\[ \Rightarrow \dfrac{x}{2} + \dfrac{x}{2} + y = 8\]
Now putting the values obtained in the formula \[AM \ge GM\], we get
\[ \Rightarrow \dfrac{{\dfrac{x}{2} + \dfrac{x}{2} + y}}{3} \ge {\left( {\dfrac{x}{2} \cdot \dfrac{x}{2} \cdot y} \right)^{\dfrac{1}{3}}}\]
Simplifying the equation further, we get
\[ \Rightarrow \dfrac{{x + y}}{3} \ge {\left( {\dfrac{{{x^2}y}}{4}} \right)^{\dfrac{1}{3}}}\]
 Rewriting the equation as
 \[ \Rightarrow {\left( {\dfrac{{{x^2}y}}{4}} \right)^{\dfrac{1}{3}}} \le \dfrac{{x + y}}{3}\]
Putting the value of \[x + y = 8\] in above inequality and we get
\[ \Rightarrow {\left( {\dfrac{{{x^2}y}}{4}} \right)^{\dfrac{1}{3}}} \le \dfrac{8}{3}\]
Taking cube in both sides of the above inequality and we get
\[ \Rightarrow {\left( {\dfrac{{{x^2}y}}{4}} \right)^{\dfrac{1}{3} \times 3}} \le {\left( {\dfrac{8}{3}} \right)^3}\]
Simplifying and we get
\[ \Rightarrow \dfrac{{{x^2}y}}{4} \le \dfrac{{{8^3}}}{{{3^3}}}\]
Simplifying further, we get
\[ \Rightarrow \dfrac{{{x^2}y}}{4} \le \dfrac{{512}}{{27}}\]
\[ \Rightarrow {x^2}y \le 4 \cdot \dfrac{{512}}{{27}}\]
So, the value comes out to be
\[ \Rightarrow {x^2}y \le \dfrac{{2048}}{{27}}\]
There the maximum value of \[{x^2}y\] is \[\dfrac{{{\bf{2048}}}}{{27}}\].
Hence, the correct option is option D.

Note: Many students make mistakes in finding value of cube of fraction. If we have a fraction \[\dfrac{p}{q}\] with \[q \ne 0\] then \[{\left( {\dfrac{p}{q}} \right)^a}\] is always written as \[\dfrac{{{p^a}}}{{{q^a}}}\] . if anyone write it as \[\dfrac{{{p^a}}}{q}\] then it gives wrong answer. We must not neglect the signs and correct formula should be used to get a valid answer.