Find the maximum and minimum of an expression

[anemone]

Find the maximum and minimum of $x^3+y^3+xy(x^2+y^2)$ given $x,\,y$ are two non-negative real numbers that satisfy $x+y+xy=3$.

 

[kaliprasad]

Find the maximum and minimum of $x^3+y^3+xy(x^2+y^2)$ given $x,\,y$ are two non-negative real numbers that satisfy $x+y+xy=3$.

not a rigorous solution but

Spoiler

we are given $x+y + xy = 3$
or $(1+x)(1+y) = 4$
we have given expression
$x^3(1+y)+y^3(1+x)$
the given condition and the expression both are symmetric in x and y so extremum occurs at $x = y = 1$
so given expression 4
other extemum is at x or y =inifinte but that makes the other value to be -ve so we have
$x=0,y=3$ or $x=3,y=0$ giving $x^3+y^3+xy(x^2+y^2)=27$
so minumum value = 4 and maximum = 27

 

[Greg]

Find the maximum and minimum of $x^3+y^3+xy(x^2+y^2)$ given $x,\,y$ are two non-negative real numbers that satisfy $x+y+xy=3$.

Spoiler

$$\begin{align*}f(x,y)=x^3+y^3+xy(x^2+y^2)&=(x+y)(x^2-xy+y^2)+xy(x+y)^2-2x^2y^2 \\
&=(x+y)[x^2-xy+y^2+xy(x+y)]-2x^2y^2 \\
&=(x+y)[x^2+y^2-xy(1-x-y)]-2x^2y^2 \\
&=(x+y)[x^2+y^2-xy(xy-2)]-2x^2y^2 \\
&=(x+y)(x^2+2xy+y^2-x^2y^2)-2x^2y^2 \\
&=(3-xy)[(3-xy)^2-x^2y^2]-2x^2y^2 \\
&=(3-xy)(9-6xy)-2x^2y^2 \\
&=27-18xy-9xy+4x^2y^2 \\
&=27-xy(27-4xy)\qquad(1)\end{align*}$$

$$x+y+xy=3\implies x=\dfrac{3-y}{y+1}$$
$$xy=\dfrac{3y-y^2}{y+1}=1-\dfrac{y^2-2y+1}{y+1}=1-\dfrac{(y-1)^2}{y+1}$$

Hence $$0\le xy\le1\qquad(2)$$

$$(1)\Leftrightarrow(2)\implies\max(f(x,y))=27$$

I don't have a proof for the minimum without using symmetry of variables.

 

[anemone]

Very well done, kaliprasad!

Spoiler

$$\begin{align*}f(x,y)=x^3+y^3+xy(x^2+y^2)&=(x+y)(x^2-xy+y^2)+xy(x+y)^2-2x^2y^2 \\
&=(x+y)[x^2-xy+y^2+xy(x+y)]-2x^2y^2 \\
&=(x+y)[x^2+y^2-xy(1-x-y)]-2x^2y^2 \\
&=(x+y)[x^2+y^2-xy(xy-2)]-2x^2y^2 \\
&=(x+y)(x^2+2xy+y^2-x^2y^2)-2x^2y^2 \\
&=(3-xy)[(3-xy)^2-x^2y^2]-2x^2y^2 \\
&=(3-xy)(9-6xy)-2x^2y^2 \\
&=27-18xy-9xy+4x^2y^2 \\
&=27-xy(27-4xy)\qquad(1)\end{align*}$$

$$x+y+xy=3\implies x=\dfrac{3-y}{y+1}$$
$$xy=\dfrac{3y-y^2}{y+1}=1-\dfrac{y^2-2y+1}{y+1}=1-\dfrac{(y-1)^2}{y+1}$$

Hence $$0\le xy\le1\qquad(2)$$

$$(1)\Leftrightarrow(2)\implies\max(f(x,y))=27$$

I don't have a proof for the minimum without using symmetry of variables.

Hi greg1313, thank you for participating and thank you for providing us the well-written solution!

Spoiler

Note that since $P=27-xy(27-4xy)$ is concave up (decreasing) function on $0\le xy\le1$, we have the minimum of $P$ at $xy=1$, so $P_{\text{minimum}}=27-1(27-4)=4$.:)

 

[kaliprasad]

Spoiler

$$\begin{align*}f(x,y)=x^3+y^3+xy(x^2+y^2)&=(x+y)(x^2-xy+y^2)+xy(x+y)^2-2x^2y^2 \\
&=(x+y)[x^2-xy+y^2+xy(x+y)]-2x^2y^2 \\
&=(x+y)[x^2+y^2-xy(1-x-y)]-2x^2y^2 \\
&=(x+y)[x^2+y^2-xy(xy-2)]-2x^2y^2 \\
&=(x+y)(x^2+2xy+y^2-x^2y^2)-2x^2y^2 \\
&=(3-xy)[(3-xy)^2-x^2y^2]-2x^2y^2 \\
&=(3-xy)(9-6xy)-2x^2y^2 \\
&=27-18xy-9xy+4x^2y^2 \\
&=27-xy(27-4xy)\qquad(1)\end{align*}$$

$$x+y+xy=3\implies x=\dfrac{3-y}{y+1}$$
$$xy=\dfrac{3y-y^2}{y+1}=1-\dfrac{y^2-2y+1}{y+1}=1-\dfrac{(y-1)^2}{y+1}$$

Hence $$0\le xy\le1\qquad(2)$$

$$(1)\Leftrightarrow(2)\implies\max(f(x,y))=27$$

I don't have a proof for the minimum without using symmetry of variables.

anemone has provided a method
here is my approach algebraically

Spoiler

$4x^2y^2-27x+27= (2xy-\dfrac{27}{2})^2 + 27 - (\dfrac{27}{2})^2$
the value is minimum when $xy = \dfrac{27}{4}$ which is > xy so as large as possible for xy which is 1