Question

If \[\dfrac{1}{x} + \dfrac{1}{y} = 1\], the function \[{a^3}{x^2} + {b^3}{y^2}\]is minimum, then x and y are
A. \[a + b,a-b\]
B. \[a,b\]
C. \[\left( {a/\left( {a + b} \right)} \right),b/\left( {a + b} \right)\]
D. \[\left( {a + b} \right)/a,\left( {a + b} \right)/b\]

Answer 1

Hint: In this question, we need to find the values of x and y. For this, first we will find the value of y from the equation \[\dfrac{1}{x} + \dfrac{1}{y} = 1\]. After that, we will find the function \[f\left( x \right)\] . We know that for minima we need to equate the derivative of \[f\left( x \right)\] to zero for getting the values of a and b.

Formula used: For finding minima, we can say that \[f'\left( x \right) = 0\]
Also, the formula of derivation is given by
\[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}\] where, u and v are differential functions of x.
And \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]

Complete step-by-step solution:
We know that \[\dfrac{1}{x} + \dfrac{1}{y} = 1\]
Let us find the value of y from the above equation.
\[\dfrac{{x + y}}{{xy}} = 1\]
Thus, we get
\[x + y = xy\]
\[x = xy - y\]
\[x = y\left( {x - 1} \right)\]
By simplifying, we get
\[y = \dfrac{x}{{x - 1}}\]
Now, put these value of x in the function \[{a^3}{x^2} + {b^3}{y^2}\]
So, we get
\[f\left( x \right) = {a^3}{x^2} + {b^3}{\left( {\dfrac{x}{{x - 1}}} \right)^2}\]
\[f\left( x \right) = {a^3}{x^2} + {b^3}\left( {\dfrac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}}}} \right)\]
Now, we will find the derivative of the above function.
 \[f'\left( x \right) = \dfrac{d}{{dx}}\left( {{a^3}{x^2} + {b^3}\left( {\dfrac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}}}} \right)} \right)\]
\[f'\left( x \right) = 2x{a^3} + \left( {\left( {{b^3}[{{\left( {x-1} \right)}^2}\;2x-{x^2}\;2\left( {x{\text{ }}-{\text{ }}1} \right)]/{{\left( {x{\text{ }}-{\text{ }}1} \right)}^4}} \right)} \right)\]
\[f'\left( x \right) = 2x{a^3} + \left( {\dfrac{{{b^3}[\left( {x-1} \right)\;2x-{x^2}\;2]}}{{{{\left( {x{\text{ }}-{\text{ }}1} \right)}^3}}}} \right)\]
By simplifying, we get
\[f'\left( x \right) = 2x{a^3} + \left( {\dfrac{{{b^3}[\;2{x^2} - 2x-2{x^2}]}}{{{{\left( {x{\text{ }}-{\text{ }}1} \right)}^3}}}} \right)\]
\[f'\left( x \right) = 2x{a^3} + \left( {\dfrac{{ - 2x{b^3}}}{{{{\left( {x{\text{ }}-{\text{ }}1} \right)}^3}}}} \right)\]
Now, put \[f'\left( x \right) = 0\]
Thus, we get
\[2x{a^3} + \left( {\dfrac{{ - 2x{b^3}}}{{{{\left( {x{\text{ }}-{\text{ }}1} \right)}^3}}}} \right) = 0\]
Let us simplify this.
\[\left( {\dfrac{{ - 2x{b^3}}}{{{{\left( {x{\text{ }}-{\text{ }}1} \right)}^3}}}} \right) = - 2x{a^3}\]
\[\left( {\dfrac{{{b^3}}}{{{{\left( {x{\text{ }}-{\text{ }}1} \right)}^3}}}} \right) = {a^3}\]
\[{b^3} = {a^3}{\left( {x{\text{ }}-{\text{ }}1} \right)^3}\]
By taking cube root on both sides, we get
\[b = a\left( {x{\text{ }}-{\text{ }}1} \right)\]
\[\dfrac{b}{a} + 1 = x\]
\[\dfrac{{a + b}}{a} = x\] …… (1)
Put this value of x in the equation \[y = \dfrac{x}{{x - 1}}\]
So, we get
\[y = \dfrac{{\dfrac{{a + b}}{a}}}{{\dfrac{{a + b}}{a} - 1}}\]
By simplifying, we get
\[y = \dfrac{{\dfrac{{a + b}}{a}}}{{\dfrac{b}{a}}}\]
\[y = \dfrac{{a + b}}{b}\]…… (2)
Thus, from (1) and (2), we can say that the values of x and y are \[\dfrac{{a + b}}{a} = x\]and \[y = \dfrac{{a + b}}{b}\] respectively.

Therefore, the correct option is (D).

Note: Many students generally make mistakes in finding the derivative of \[f\left( x \right)\] , specifically while finding the derivative of a fraction. Students may confuse with the middle sign. Also, they may get the wrong end result as the derivative of the function gets wrong.