MHB Find the maximum and minimum of an expression

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The discussion focuses on finding the maximum and minimum values of the expression \(x^3+y^3+xy(x^2+y^2)\) under the constraint \(x+y+xy=3\) for non-negative real numbers \(x\) and \(y\). Participants share various approaches, including algebraic methods and solutions provided by others. The conversation highlights the collaborative nature of problem-solving in mathematics, with users expressing appreciation for well-formulated solutions. Overall, the thread emphasizes the importance of exploring different methods to tackle mathematical expressions constrained by specific conditions. The goal remains to identify the optimal values of the given expression.
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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 $x+y+xy=3$.
 
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anemone said:
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
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
 
anemone said:
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 $x+y+xy=3$.

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

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

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.
 
Very well done, kaliprasad!:cool:

greg1313 said:
$$\begin{align*}f(x,y)=x^3+y^3+xy(x^2+y^2)&=(x+y)(x^2-xy+y^2)+xy(x+y)^2-2x^2y^2 \\
&=(x+y)[x^2-xy+y^2+xy(x+y)]-2x^2y^2 \\
&=(x+y)[x^2+y^2-xy(1-x-y)]-2x^2y^2 \\
&=(x+y)[x^2+y^2-xy(xy-2)]-2x^2y^2 \\
&=(x+y)(x^2+2xy+y^2-x^2y^2)-2x^2y^2 \\
&=(3-xy)[(3-xy)^2-x^2y^2]-2x^2y^2 \\
&=(3-xy)(9-6xy)-2x^2y^2 \\
&=27-18xy-9xy+4x^2y^2 \\
&=27-xy(27-4xy)\qquad(1)\end{align*}$$

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

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!
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 $P_{\text{minimum}}=27-1(27-4)=4$.:)
 
Last edited:
greg1313 said:
$$\begin{align*}f(x,y)=x^3+y^3+xy(x^2+y^2)&=(x+y)(x^2-xy+y^2)+xy(x+y)^2-2x^2y^2 \\
&=(x+y)[x^2-xy+y^2+xy(x+y)]-2x^2y^2 \\
&=(x+y)[x^2+y^2-xy(1-x-y)]-2x^2y^2 \\
&=(x+y)[x^2+y^2-xy(xy-2)]-2x^2y^2 \\
&=(x+y)(x^2+2xy+y^2-x^2y^2)-2x^2y^2 \\
&=(3-xy)[(3-xy)^2-x^2y^2]-2x^2y^2 \\
&=(3-xy)(9-6xy)-2x^2y^2 \\
&=27-18xy-9xy+4x^2y^2 \\
&=27-xy(27-4xy)\qquad(1)\end{align*}$$

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

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

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