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

How do you minimize and maximize f(x,y)=x−yx2f\left( x,y \right)=\dfrac{x-y}{{{x}^{2}}} constrained to xy=4xy=4?

Explanation

Solution

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=4xy=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(x,y)=x−yx2f\left( x,y \right)=\dfrac{x-y}{{{x}^{2}}} Constrained to xy=4xy=4.
Converting into single variable calculus.
We have, xy=4xy=4, so y=4xy=\dfrac{4}{x}
Therefore, putting y=4xy=\dfrac{4}{x} inf(x,y)=x−yx2f\left( x,y \right)=\dfrac{x-y}{{{x}^{2}}}, we get
f(x)=x−4x(x)2=1x−4x3f\left( x \right)=\dfrac{x-\dfrac{4}{x}}{{{\left( x \right)}^{2}}}=\dfrac{1}{x}-\dfrac{4}{{{x}^{3}}}
f(x)=1x−4x3f\left( x \right)=\dfrac{1}{x}-\dfrac{4}{{{x}^{3}}}
Differentiate f(x), we get
f′(x)=−1x2+12x4f'\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
−1x2+12x4=0-\dfrac{1}{{{x}^{2}}}+\dfrac{12}{{{x}^{4}}}=0
Simplifying the above expression by taking the LCM of denominator, we get
−x2x4+12x4=0-\dfrac{{{x}^{2}}}{{{x}^{4}}}+\dfrac{12}{{{x}^{4}}}=0
Thus,
12−x2=012-{{x}^{2}}=0
Solving for the value of ‘x’, we get
x=±23x=\pm 2\sqrt{3}
Now, putting the value of x=±23x=\pm 2\sqrt{3} in y=4xy=\dfrac{4}{x}, we get
y=±423=±23y=\pm \dfrac{4}{2\sqrt{3}}=\pm \dfrac{2}{\sqrt{3}}
So, therefore we get two fixed point, i.e.
⇒(23,23),(−23,−23)\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′(x)=−1x2+12x4f'\left( x \right)=-\dfrac{1}{{{x}^{2}}}+\dfrac{12}{{{x}^{4}}}
f′′(x)=2x3−48x5f''\left( x \right)=\dfrac{2}{{{x}^{3}}}-\dfrac{48}{{{x}^{5}}}
Taking out 2x3\dfrac{2}{{{x}^{3}}}as a common factor, we get
f′′(x)=2x3(1−24x2)f''\left( x \right)=\dfrac{2}{{{x}^{3}}}\left( 1-\dfrac{24}{{{x}^{2}}} \right)
Putting x=23x=2\sqrt{3} in the second derivative we get
f′′(x)=2(23)3(1−24(23)2)=2(23)3(1−2412)=2(23)3(1−2)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=23x=2\sqrt{3} in second derivative, we will get negative value.
So this is a local maximum.
Putting x=−23x=-2\sqrt{3} in the second derivative we get
f′′(x)=−2(23)3(1−24(−23)2)=−2(23)3(1−2412)=−2(23)3(1−2)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=−23x=-2\sqrt{3} in second derivative, we will get positive value.
So this is a local minimum.
Now,
Function is maximum at(23,23)\left( 2\sqrt{3},\dfrac{2}{\sqrt{3}} \right).
Putting (x, y) = (23,23)\left( 2\sqrt{3},\dfrac{2}{\sqrt{3}} \right) in f(x, y), we get
⇒f(23,23)=23−23(23)2=43×112=133\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(−23,−23)\left( -2\sqrt{3},-\dfrac{2}{\sqrt{3}} \right).
Putting (x, y) = (−23,−23)\left( -2\sqrt{3},-\dfrac{2}{\sqrt{3}} \right) in f(x, y), we get
⇒f(−23,−23)=−23−(−23)(−23)2=−43×112=−133\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,
fmax⁡=f(23,23)=133{{f}_{\max }}=f\left( 2\sqrt{3},\dfrac{2}{\sqrt{3}} \right)=\dfrac{1}{3\sqrt{3}}
fmin⁡=f(−23,−23)=−133{{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.