# 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
5 years, 11 months ago

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A much easier and convenient way would be this:

$x^2-y^2=6-2xy$

On squaring we have:

$x^4+y^4=36+6(xy)^2-24xy$

Let $N=(x^2+y^2)^2$ for some $N \in \mathbb{R}$ obviously $N \geq 0$

$N=x^4+y^4+2(xy)^2$

$N=36+8(xy)^2-24xy$

Substitute the value of $x^4+y^4=36+6(xy)^2-24xy$

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

On factorising:

$\large N=8((xy-\frac{3}{2})^2 -\frac{9}{2} + \frac{9}{4})$

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

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

We get the minimum to be $18$.

- 5 years, 11 months ago

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

Staff - 5 years, 11 months ago

Yes I was about to do that in the edit.

- 5 years, 11 months ago

Can't it be solved using trigonometry?

- 5 years, 11 months ago

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

- 5 years, 11 months ago

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

- 5 years, 11 months ago

Differentiate the first equation with respect to $x$.

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

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

$\large \frac{dy}{dx}=\frac{x+y}{y-x}$

Similarly differentiating the second equation with respect to $x$:

$\large \frac{dN}{dx}=2(x^2+y^2)(2x+2y\frac{dy}{dx})$

In order to minimize $N$ we have $\frac{dN}{dx}=0$

As such we have:

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

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

So we have that:

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

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

$y^2-x^2+2xy=0$

Now, $y^2=2xy+x^2-6$

$\Rightarrow 2xy+x^2-6-x^2+2xy=0$

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

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

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

$N=4(x^2+xy-3)^2$

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

$x^2-y^2=6-3$

Squaring both sides gives us:

$x^4+y^4-2(xy)^2=9$

$x^4+y^4=\frac{27}{2}$

Now,

$N=(x^4+y^4+2(xy)^2$

$\Rightarrow N=\frac{27}{2}+2\frac{9}{4}$

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

- 5 years, 11 months ago