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Find minimum value of xy

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
Hi MHB,

I've found an interesting and difficult (at least for me) problem which asks us to find the minimum value of $xy$ and no matter how serious that I tried, I couldn't get myself out of the misery of finding what the question wanted. Could anyone help?

Problem:

Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hi MHB,

I've found an interesting and difficult (at least for me) problem which asks us to find the minimum value of $xy$ and no matter how serious that I tried, I couldn't get myself out of the misery of finding what the question wanted. Could anyone help?

Problem:

Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.
It is a classical problem of the search of the minimum of a function $\displaystyle f(x,y)$ under the condition $\displaystyle g(x,y)=0$. In Your case is $\displaystyle f(x,y)= x\ y$ and g(*,*) is derived solving the system...


$\displaystyle x^2+y^2+z^2=7,\ xy+xz+yz=4\ \implies g(x,y)= x^{2} + y^{2} + \frac{(4 - x\ y)^{2}}{(x+y)^{2}} - 7\ (1)$

The standard approach uses Lagrange multipliers...

Kind regards


$\chi$ $\sigma$
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Perhaps could be useful an algebraic approach. Using simultaneous diagonalization of quadratic forms, there exists an orthogonal matrix $P\in\mathbb{R}^{3\times 3}$ such that $X=PU$ with $X=(x,y,z)^T$ and $U=(u,v,w)^T$ transforms $D\equiv\left \{ \begin{matrix} x^2+y^2+z^2=7 \\xy+xz+yz=4\end{matrix}\right.$ into $D'\equiv\left \{ \begin{matrix} u^2+v^2+w^2=7 \\2u^2-v^2-w^2=8\end{matrix}\right.$ so, necessarily $u^2=15,$ etc.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.
First, the minimum value of $xy$ must be positive, because if $xy\leqslant 0$ then $(x+y)z\geqslant 4$. So $(x+y)^2\geqslant\dfrac{16}{z^2}$, and $$7-z^2 = x^2+y^2 = (x+y)^2 - 2xy \geqslant (x+y)^2 \geqslant \frac{16}{z^2}.$$ Thus $z^2 + \dfrac{16}{z^2} \leqslant 7$. But that cannot happen, because the minimum value of $z^2 + \dfrac{16}{z^2}$ is $8$ (occurring when $z^2 = 4$).

So we may assume that $xy>0$. Let $u = \sqrt{xy}$ and $v = x+y$. Then we can write the equations as $v^2-2u^2 + z^2 = 7$, $vz+u^2=4$. Therefore $v^2 - 2(4-vz) + z^2 = 7,$ so $(v+z)^2 = 15$. But $vz\leqslant\bigl(\frac12(v+z)\bigr)^2 = \frac{15}4.$ Therefore $xy = u^2 \geqslant 4-\frac{15}4 = \frac14$.

To see that the minimum value $xy=\frac14$ is attained, check that if $x= \frac14(\sqrt{15} + \sqrt{11})$, $y= \frac14(\sqrt{15} - \sqrt{11})$ and $z= \frac12\sqrt{15}$, then the original equations are satisfied, and $xy = \frac14$.

To obtain those values for $x$, $y$ and $z$, notice that the previous equations imply that to attain the minimum value $\frac14$ for $u$, we must have $v=z=\frac12\sqrt{15}$. In terms of $x$ and $y$, this says that $x+y = \frac12\sqrt{15}$. Also, $xy = \frac14$. This means that $x$ and $y$ are the roots of the quadratic equation $\lambda^2 - \frac12\sqrt{15}\lambda + \frac14 = 0$, giving the values for $x$ and $y$ listed above.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Let me show you how the method of the Lagrange Multipliers would work.
You will probably learn it sooner or later anyway and it may interest you.

We'll define the function $F(x,y,z,\lambda,\mu)$ as follows:
$$F(x,y,z,\lambda,\mu) = xy + \lambda(x^2+y^2+z^2-7) + \mu(xy+xz+yz-4)$$

What you want to find is the extremum of $F$.
To find it, you need to calculate each of the partial derivatives and set each of them to zero.
That is:
\begin{array}{lclcl}
\frac{\partial F}{\partial x} &=& y + 2\lambda x + \mu(y+z) &=& 0 \\
\frac{\partial F}{\partial y} &=& x + 2\lambda y + \mu(x+z) &=& 0 \\
\frac{\partial F}{\partial z} &=& \phantom{x + } 2\lambda z + \mu(x+y) &=& 0 \\
\frac{\partial F}{\partial \lambda} &=& x^2+y^2+z^2-7 &=& 0 \\
\frac{\partial F}{\partial \mu} &=& xy + xz + yz - 4 &=& 0 \\
\end{array}

The last 2 equations may look familiar. ;)

Anyway, this is a system of 5 equations with 5 variables.
Solve it, at least for x, y, and z, and you have your extremum.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
Hi chisigma, and Fernando Revilla, thanks for the helpful replies, I do appreciate them!

Hi Opalg, I too approached it the way you did and I too used the AM-GM inequality for the part $\dfrac{v+z}{2}\ge \sqrt{vz}$, but what I didn't do is to let $u=\sqrt{vz}$ and let $v=xy$ and what I obtained was two messy equations which I just wanted to abandon and ask for help here at MHB...Opalg, you know, you're one of my heroes in the field of mathematics! Thank you for everything!

Hi I like Serena, I always ask MarkFL to teach me how to approach optimization questions using the Lagrange Multipliers and in fact, he did teach me, but perhaps I am too lazy to learn. I never seemed to grasp this concept well but now you have raised it to my attention, I feel it's high time for me to pick up the concept...also, I have never seen Opalg used Lagrange Multipliers in this forum before, so I figured maybe is not a very handy method...hehehe...and I want to thank you for your kindness to want to teach me LM too!
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
I have never seen Opalg used Lagrange Multipliers in this forum before, so I figured maybe is not a very handy method...hehehe...
That does not mean that I don't value the Lagrange multiplier method. It is a very powerful tool. But if a problem seems algebraic in nature then I prefer to tackle it with algebraic methods rather than resorting to calculus. I find that an algebraic or geometric solution often gives more insight into the nature of a problem than an analytic solution (and that is despite being an analyst myself).
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
That does not mean that I don't value the Lagrange multiplier method. It is a very powerful tool. But if a problem seems algebraic in nature then I prefer to tackle it with algebraic methods rather than resorting to calculus. I find that an algebraic or geometric solution often gives more insight into the nature of a problem than an analytic solution (and that is despite being an analyst myself).
I'm sorry Opalg...for making a statement like you don't value Lagrange multiplier...I wish I could take back what I said in my previous reply and I will be more careful from now on and I promise it won't happen again.