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

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TopNewestA 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 \(18\). – Aditya Parson · 4 years ago

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– Calvin Lin Staff · 4 years 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.Log in to reply

– Kiran Patel · 4 years ago

Can't it be solved using trigonometry?Log in to reply

– Aditya Parson · 4 years ago

Yes I was about to do that in the edit.Log in to reply

– Kiran Patel · 4 years ago

I think \(N\geq0\) is incorrect.It must be N>0.Log in to reply

– Aditya Parson · 4 years 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.Log in to reply

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. – Aditya Parson · 4 years ago

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