# 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, 6 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, 6 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, 6 months ago

Yes I was about to do that in the edit.

- 5 years, 6 months ago

Can't it be solved using trigonometry?

- 5 years, 6 months ago

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

- 5 years, 6 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, 6 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, 6 months ago