Find the minimum value!

\[If{\kern 1pt} {x^2}{\kern 1pt} + {\kern 1pt} 2xy{\kern 1pt} - {\kern 1pt} {y^2} = {\kern 1pt} 6.Then{\kern 1pt} {\kern 1pt} find{\kern 1pt} {\kern 1pt} the{\kern 1pt} {\kern 1pt} minimum{\kern 1pt} {\kern 1pt} value{\kern 1pt} {\kern 1pt} of{\kern 1pt} {({x^2} + {y^2})^2}?\] where x and y are real numbers.

Note by Kiran Patel
7 years, 11 months ago

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Comments

A much easier and convenient way would be this:

x2−y2=6−2xyx^2-y^2=6-2xy

On squaring we have:

x4+y4=36+6(xy)2−24xyx^4+y^4=36+6(xy)^2-24xy

Let N=(x2+y2)2N=(x^2+y^2)^2 for some N∈RN \in \mathbb{R} obviously N≥0N \geq 0

N=x4+y4+2(xy)2N=x^4+y^4+2(xy)^2

N=36+8(xy)2−24xyN=36+8(xy)^2-24xy

Substitute the value of x4+y4=36+6(xy)2−24xyx^4+y^4=36+6(xy)^2-24xy

N=8((xy)2−3xy+92)\large N=8((xy)^2-3xy+\frac{9}{2})

On factorising:

N=8((xy−32)2−92+94)\large N=8((xy-\frac{3}{2})^2 -\frac{9}{2} + \frac{9}{4})

Clearly the minimum occurs at xy=32xy=\frac{3}{2}

⇒N=18\Rightarrow N=18 at xy=32xy=\frac{3}{2} where (x,y)=(18+32,18−32)(x, y)=(\large \sqrt {\frac {\sqrt{18}+3}{2}},\sqrt {\frac {\sqrt{18}-3}{2}})

We get the minimum to be 1818.

Aditya Parson - 7 years, 11 months ago

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As with all inequality questions, you need to verify that equality can actually hold. Simply stating that xy=32xy = \frac{3}{2} is not sufficient to guarantee that real values of xx and yy exist. E.g. you could have complex solutions to the equation.

Calvin Lin Staff - 7 years, 11 months ago

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Yes I was about to do that in the edit.

Aditya Parson - 7 years, 11 months ago

Can't it be solved using trigonometry?

Kiran Patel - 7 years, 11 months ago

I think N≥0N\geq0 is incorrect.It must be N>0.

Kiran Patel - 7 years, 11 months ago

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Yes, if you take into consideration the first equation N>0N>0 is more accurate. I stated that N≥0N \geq 0 without considering the first equation.

Aditya Parson - 7 years, 11 months ago

Differentiate the first equation with respect to xx.

We get: Let N=(x2+y2)2N=(x^2+y^2)^2

2x+2y+2x(dydx)−2y(dydx)=0\large 2x+2y+2x(\frac{dy}{dx})-2y(\frac{dy}{dx})=0

dydx=x+yy−x\large \frac{dy}{dx}=\frac{x+y}{y-x}

Similarly differentiating the second equation with respect to xx:

dNdx=2(x2+y2)(2x+2ydydx)\large \frac{dN}{dx}=2(x^2+y^2)(2x+2y\frac{dy}{dx})

In order to minimize NN we have dNdx=0\frac{dN}{dx}=0

As such we have:

x2+y2=0x^2+y^2=0 or (2x+2ydydx)=0 (2x+2y\frac{dy}{dx})=0

Note that first of the above equation gives us x=−y2x=\sqrt{-y^2} which is not possible since x,y∈Rx,y \in \mathbb{R} .

So we have that:

2x+2ydydx=0\large 2x+2y\frac{dy}{dx}=0

Substituting for dydx\frac{dy}{dx}, gives us:

y2−x2+2xy=0y^2-x^2+2xy=0

Now, y2=2xy+x2−6y^2=2xy+x^2-6

⇒2xy+x2−6−x2+2xy=0\Rightarrow 2xy+x^2-6-x^2+2xy=0

xy=32xy=\frac{3}{2}

This can be verified to be the minimum value for xyxy by checking the sign of d2ydx2\frac{d^2y}{dx^2} or maybe we could just use the fact that since it is obvious that NN will never reach a maximum so xy=32xy=\frac{3}{2} will give the minimum.(I am not exactly sure)

We can re-write N=(x2+x2+2xy−6)2\large N=(x^2+x^2+2xy-6)^2

N=4(x2+xy−3)2N=4(x^2+xy-3)^2

Since we have xy=32xy=\frac {3}{2}

x2−y2=6−3x^2-y^2=6-3

Squaring both sides gives us:

x4+y4−2(xy)2=9x^4+y^4-2(xy)^2=9

x4+y4=272x^4+y^4=\frac{27}{2}

Now,

N=(x4+y4+2(xy)2N=(x^4+y^4+2(xy)^2

⇒N=272+294\Rightarrow N=\frac{27}{2}+2\frac{9}{4}

N=18N=18 is the minimum value of the expression.

Aditya Parson - 7 years, 11 months ago
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