Question

How do you minimize and maximize \[f\left( x,y \right)=\dfrac{x-y}{{{x}^{2}}}\] constrained to \[xy=4\]?

Answer 1

Hint: In the given question, we have been asked to find the maxima and the minima of a function and it is given that the function is constrained to \[xy=4\]. In order to find the maxima and minima first we need to convert the function into single variable calculus. Then finding the first derivative of a single variable calculus function and equation it with 0, we will get two fixed points. Later we will do a second derivative for the similar function and checking whether the value is positive and negative.

Complete step by step solution:
We have given that,
\[f\left( x,y \right)=\dfrac{x-y}{{{x}^{2}}}\] Constrained to \[xy=4\].
Converting into single variable calculus.
We have, \[xy=4\], so \[y=\dfrac{4}{x}\]
Therefore, putting \[y=\dfrac{4}{x}\] in\[f\left( x,y \right)=\dfrac{x-y}{{{x}^{2}}}\], we get
\[f\left( x \right)=\dfrac{x-\dfrac{4}{x}}{{{\left( x \right)}^{2}}}=\dfrac{1}{x}-\dfrac{4}{{{x}^{3}}}\]
\[f\left( x \right)=\dfrac{1}{x}-\dfrac{4}{{{x}^{3}}}\]
Differentiate f(x), we get
\[f'\left( x \right)=-\dfrac{1}{{{x}^{2}}}+\dfrac{12}{{{x}^{4}}}\]
Solving the above expression, we get
Putting f’(x) equals to 0, we get
\[-\dfrac{1}{{{x}^{2}}}+\dfrac{12}{{{x}^{4}}}=0\]
Simplifying the above expression by taking the LCM of denominator, we get
\[-\dfrac{{{x}^{2}}}{{{x}^{4}}}+\dfrac{12}{{{x}^{4}}}=0\]
Thus,
\[12-{{x}^{2}}=0\]
Solving for the value of ‘x’, we get
\[x=\pm 2\sqrt{3}\]
Now, putting the value of \[x=\pm 2\sqrt{3}\] in \[y=\dfrac{4}{x}\], we get
\[y=\pm \dfrac{4}{2\sqrt{3}}=\pm \dfrac{2}{\sqrt{3}}\]
So, therefore we get two fixed point, i.e.
\[\Rightarrow \left( 2\sqrt{3},\dfrac{2}{\sqrt{3}} \right),\left( -2\sqrt{3},-\dfrac{2}{\sqrt{3}} \right)\]
Second derivation of f(x), we get
\[f'\left( x \right)=-\dfrac{1}{{{x}^{2}}}+\dfrac{12}{{{x}^{4}}}\]
\[f''\left( x \right)=\dfrac{2}{{{x}^{3}}}-\dfrac{48}{{{x}^{5}}}\]
Taking out \[\dfrac{2}{{{x}^{3}}}\]as a common factor, we get
\[f''\left( x \right)=\dfrac{2}{{{x}^{3}}}\left( 1-\dfrac{24}{{{x}^{2}}} \right)\]
Putting \[x=2\sqrt{3}\] in the second derivative we get
\[f''\left( x \right)=\dfrac{2}{{{\left( 2\sqrt{3} \right)}^{3}}}\left( 1-\dfrac{24}{{{\left( 2\sqrt{3} \right)}^{2}}} \right)=\dfrac{2}{{{\left( 2\sqrt{3} \right)}^{3}}}\left( 1-\dfrac{24}{12} \right)=\dfrac{2}{{{\left( 2\sqrt{3} \right)}^{3}}}\left( 1-2 \right)\]
, i.e. when we putting the value of \[x=2\sqrt{3}\] in second derivative, we will get negative value.
So this is a local maximum.
Putting \[x=-2\sqrt{3}\] in the second derivative we get
\[f''\left( x \right)=-\dfrac{2}{{{\left( 2\sqrt{3} \right)}^{3}}}\left( 1-\dfrac{24}{{{\left( -2\sqrt{3} \right)}^{2}}} \right)=-\dfrac{2}{{{\left( 2\sqrt{3} \right)}^{3}}}\left( 1-\dfrac{24}{12} \right)=-\dfrac{2}{{{\left( 2\sqrt{3} \right)}^{3}}}\left( 1-2 \right)\], i.e. when we putting the value of \[x=-2\sqrt{3}\] in second derivative, we will get positive value.
So this is a local minimum.
Now,
Function is maximum at\[\left( 2\sqrt{3},\dfrac{2}{\sqrt{3}} \right)\].
Putting (x, y) = \[\left( 2\sqrt{3},\dfrac{2}{\sqrt{3}} \right)\] in f(x, y), we get
\[\Rightarrow f\left( 2\sqrt{3},\dfrac{2}{\sqrt{3}} \right)=\dfrac{2\sqrt{3}-\dfrac{2}{\sqrt{3}}}{{{\left( 2\sqrt{3} \right)}^{2}}}=\dfrac{4}{\sqrt{3}}\times \dfrac{1}{12}=\dfrac{1}{3\sqrt{3}}\]
Function is minimum at\[\left( -2\sqrt{3},-\dfrac{2}{\sqrt{3}} \right)\].
Putting (x, y) = \[\left( -2\sqrt{3},-\dfrac{2}{\sqrt{3}} \right)\] in f(x, y), we get
\[\Rightarrow f\left( -2\sqrt{3},-\dfrac{2}{\sqrt{3}} \right)=\dfrac{-2\sqrt{3}-\left( -\dfrac{2}{\sqrt{3}} \right)}{{{\left( -2\sqrt{3} \right)}^{2}}}=-\dfrac{4}{\sqrt{3}}\times \dfrac{1}{12}=-\dfrac{1}{3\sqrt{3}}\]

Therefore,
\[{{f}_{\max }}=f\left( 2\sqrt{3},\dfrac{2}{\sqrt{3}} \right)=\dfrac{1}{3\sqrt{3}}\]
\[{{f}_{\min }}=f\left( -2\sqrt{3},-\dfrac{2}{\sqrt{3}} \right)=-\dfrac{1}{3\sqrt{3}}\]
Hence, it is the required answer.

Note: Students should know about the concept of finding the maximum and the minimum of the function i.e. first we need to find f’(x)=0 and the solution of the equation, then solve for f’’(x) < 0. Maxima and minima of any function is known as the largest and the smallest value of a given function with the constraint range that is given or on the whole range.