Matematické Fórum

stuart clark · 08. 12. 2014 04:43

Finding $(a){\max\limits_{x,y\in [0,1]}(x^2y-xy^2)}$ and $(b){\max\limits_{x,y,z\in [0,1]}(x^2y+y^2z+z^2x-xy^2-yz^2-zx^2)}$

Pavel · 15. 01. 2015 09:44

↑ stuart clark:

(a) Let us define a quadratic function $Q(y):=x^2y-xy^2$ of a single variable which depends on a fixed real parameter $x$ . Then

$Q(y):=x^2y-xy^2=-xy(y-x)$

and it is easy to see, that $y=0$ and $y=x$ are two roots of $Q(y)$ . Thus $Q(y)$ has a global maximum point $y^*=\frac x2$ , so the maximal value of $Q(y)$ is

$Q(y^*)=x^2\cdot\frac x2-x\left(\frac x2\right)^2=\frac{x^3}4.$

Since $x\in[0,1]$ , we have

${\color{blue}\max\limits_{x,y\in [0,1]}(x^2y-xy^2)=\max\limits_{x\in [0,1]}Q(y^*)=\frac 14.}$

The equality is satisfied for ${\color{blue}x=1}$ and ${\color{blue}y=\frac 12}$

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(b) We use the same technique.

Let us define $Q(x):=x^2y+y^2z+z^2x-xy^2-yz^2-zx^2=-x^2(z-y)+x(z^2-y^2)+y^2z-yz^2$ as a quadratic function of a single variable $x$ depending on two fixed real parameters $y,z$ . It has two roots $x=y$ and $x=z$ , so it has a global maximum point $x^*=\frac{y+z}{2}$ . Maximal value of $Q(x)$ is

$Q(x^*)&=\frac 14(y+z)^2(y-z)+\frac 12(y+z)(z^2-y^2)+y^2z-yz^2\\ &=\frac 14(y+z)^2(y-z)-\frac 12(y+z)^2(y-z)+yz(y-z)\\ &=-\frac 14(y+z)^2(y-z)+\left(\frac{(y+z)^2-(y-z)^2}4\right)(y-z)\\ &=-\frac 14(y+z)^2(y-z)+\frac 14(y+z)^2(y-z)-\frac 14(y-z)^3\\ &=\frac 14(z-y)^3.$

Since $y,z\in[0,1]$ , $Q(x^*)$ is maximal, if ${\color{blue}z=1}$ and ${\color{blue}y=0}$ . Thus ${\color{blue}x=\frac 12}$ and

${\color{blue}\max\limits_{x,y,z\in [0,1]}(x^2y+y^2z+z^2x-xy^2-yz^2-zx^2)=\frac 14.}$

If we denote $P(x,y,z):=x^2y+y^2z+z^2x-xy^2-yz^2-zx^2$ then obviously

$P(x,y,z)=P(y,z,x)=P(z,x,y).$

The given identity is satisfied also for ${\color{blue}(x,y,z)=\left(0,1,\frac 12\right)}$ and ${\color{blue}(x,y,z)=\left(1,\frac 12,0\right)}$ .

vlado_bb · 15. 01. 2015 10:15

↑ stuart clark:Elemenary solution of a:
1. $Q$ is symmetric w.r.t. the line $y=x$ , hence it is sufficient to work on the lower triangle.
2. $Q$ is nondecreasing along the lines $y=kx$ on this triangle, hence the maximum will be of the type $[1,y]$ .
3. The rest is trivial.

Matematické Fórum

#1 08. 12. 2014 04:43

maximum of f(x,y)

#2 15. 01. 2015 09:44

Re: maximum of f(x,y)

#3 15. 01. 2015 10:15

Re: maximum of f(x,y)

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