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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 $x+y+xy=3$.
not a rigorous solution butanemone 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 $x+y+xy=3$.
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 $x+y+xy=3$.
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.
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.
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.
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.
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.
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.
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.