Find the maximum and minimum of an expression

We can rewrite the given equation as $x+y+xy=3$ and manipulate it to get $(x+1)(y+1)=4$. Using this, we can substitute $y=\frac{4}{x+1}$ into the expression $x^3+y^3+xy(x^2+y^2)$. Simplifying, we get $x^3+\frac{64}{(x+1)^3}+\frac{4x^3}{x+1}$. Taking the derivative and setting it equal to 0, we get $x^3=-\frac{16}{x+1}$. Solving, we get $x=-1, 2$. Plugging in these values into
  • #1
anemone
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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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  • #2
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
 
  • #3
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 $x+y+xy=3$.

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

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

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

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

I don't have a proof for the minimum without using symmetry of variables.
 
  • #4
Very well done, kaliprasad!:cool:

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

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

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

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

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:
  • #5
greg1313 said:
\(\displaystyle \begin{align*}f(x,y)=x^3+y^3+xy(x^2+y^2)&=(x+y)(x^2-xy+y^2)+xy(x+y)^2-2x^2y^2 \\
&=(x+y)[x^2-xy+y^2+xy(x+y)]-2x^2y^2 \\
&=(x+y)[x^2+y^2-xy(1-x-y)]-2x^2y^2 \\
&=(x+y)[x^2+y^2-xy(xy-2)]-2x^2y^2 \\
&=(x+y)(x^2+2xy+y^2-x^2y^2)-2x^2y^2 \\
&=(3-xy)[(3-xy)^2-x^2y^2]-2x^2y^2 \\
&=(3-xy)(9-6xy)-2x^2y^2 \\
&=27-18xy-9xy+4x^2y^2 \\
&=27-xy(27-4xy)\qquad(1)\end{align*}\)

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

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

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

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
 

FAQ: Find the maximum and minimum of an expression

What is the purpose of finding the maximum and minimum of an expression?

The purpose of finding the maximum and minimum of an expression is to determine the highest and lowest possible values that the expression can take on. This information can be useful in various applications, such as optimization problems or analyzing data sets.

How do you find the maximum and minimum of an expression?

To find the maximum and minimum of an expression, you can use techniques such as differentiation, completing the square, or graphing. These methods involve finding critical points, where the derivative of the expression is equal to zero, and evaluating the expression at these points to determine the maximum and minimum values.

What is a critical point and how does it relate to finding the maximum and minimum of an expression?

A critical point is a point where the derivative of an expression is equal to zero. This means that the slope of the graph of the expression is flat at that point. Critical points are important in finding the maximum and minimum of an expression because the maximum and minimum values occur at these points.

Can an expression have more than one maximum or minimum value?

Yes, an expression can have multiple maximum or minimum values. This can happen when the graph of the expression has multiple peaks or valleys. In this case, there will be multiple critical points where the derivative is equal to zero, and the maximum and minimum values will occur at these points.

How can finding the maximum and minimum of an expression be applied in real life?

Finding the maximum and minimum of an expression can be applied in various real-life scenarios, such as determining the most profitable production level, optimizing resource allocation, or analyzing market trends. It can also be used in engineering and scientific fields to determine the optimal design or solution to a problem.

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