Find minimum value of the function If $g(x) = \max|y^2 - xy|$ for $0\leq y\leq 1$. Then the minimum value of $g(x)$ is?
I am not being able to proceed. Tried drawing the graph.
 A: Computing the function $g(x) = \max{|y^2 - xy|}$ is equivalent to solving for the maximum value of the expression $|y^2 - xy| = f(y)$ with a fixed parameter $x$. Let's then imagine that $x$ is a constant. In general, the maximum value of a function (of one variable) can be found in either


*

*Where the derivative is zero;

*Where the derivative is not defined; or

*At the boundaries of the domain. 


Let's go through these cases one by one. Firstly, we want to find $\frac{d f(y)}{dy}=0$. For $y^2 - xy < 0$ the result is $y = \frac{x}{2}$, and for $y^2 -xy > 0$ there is no solution. 
The derivative is not defined when $y^2 - xy = 0 \Rightarrow y= 0$ or $y = x$.
The boundaries of the domain were defined to be $y = 0$ and $y = 1$. 
Inserting these results into $f(y)$ gives us $f(\frac{x}{2}) = \frac{x^2}{4}$; $f(0)=f(x)=0$; $f(1)=|1-x|$. We need the largest of these, which is
$g(x) = \begin{cases}
\left| 1 - x \right| , \text{if} -2(1+\sqrt{2}) \leq x \leq 2(\sqrt{2}-1)\\
x^2/4 , \text{otherwise}
\end{cases}$
This was found out by solving the equation $|1-x|>\frac{x^2}{4}$.
The minimum of $g(x)$ can be found at $x= 2(\sqrt{2}-1)$, which is $g(2(\sqrt{2}-1)) = 1- 2(\sqrt{2}-1) = 3-2\sqrt{2}$.
A: The maximum for the square of a non negative function is the square of the maximum , so find the maximum (x fixed) for  z= |...|^2=( y^2-xy)^2 = y^2(y-x)^2  this gets rid of the troublesome absolute value . Now study z as a function of y for fixed x . Determine where dz/dx is =,-,0 and sketch the graph .
A: \begin{align*}
&\text{First suppose $x < 0$.}\\[6pt]
&\text{Then}\;\;x < 0\\[4pt]
&\implies\, y > x&&\text{[since $0 \le y \le 1$]}\\[4pt]
&\implies\, y(y-x) \ge 0\\[4pt]
&\implies\, y^2-xy \ge 0\\[4pt]
&\implies\, g(x) = \max(y^2-xy)\\[4pt]
&\qquad\qquad\qquad\;\;\text{(for $0 \le y \le 1$)}\\[4pt]
&\implies\, g(x) = \max(0,1-x)&&\text{[no local max, so}\\[2pt]
&&&\,\text{max must occur at $y=0$ or $y=1$]}\\[4pt]
&\implies\, g(x) = 1-x\\[12pt]
&\text{Next suppose $1 < x \le 2$.}\\[6pt]
&\text{Then}\;\;1 < x \le 2\\[4pt]
&\implies\, y < x&&\text{[since $0 \le y \le 1$]}\\[4pt]
&\implies\, y(x-y) \ge 0\\[4pt]
&\implies\, xy-y^2\ge 0\\[4pt]
&\implies\, g(x) = \max(xy-y^2)\\[2pt]
&\qquad\qquad\qquad\;\;\text{(for $0 \le y \le 1$)}\\[4pt]
&\implies\, g(x) = \max(0,x-1,x^2/4)&&\text{[max occurs at}\\[2pt]
&&&\,\text{$y=0,\;y=1$}\\[2pt]
&&&\,\text{or$\;y=x/2$ (critical point)]}\\[6pt]
&\implies\, g(x) = x^2/4&&\text{[since $x^2/4 \ge x-1$]}\\[12pt]
&\text{Next suppose $x > 2$.}\\[6pt]
&\text{Then}\;\;x>2\\[4pt]
&\implies\, y < x&&\text{[since $0 \le y \le 1$]}\\[4pt]
&\implies\, y(x-y) \ge 0\\[4pt]
&\implies\, xy-y^2\ge 0\\[4pt]
&\implies\, g(x) = \max(xy-y^2)\\[4pt]
&\implies\, g(x) = \max(0,x-1)&&\text{[max occurs at}\\[4pt]
&&&\,\text{$y=0,\;y=1$}\\[4pt]
&&&\,\text{($y = x/2 \notin [0,1]$)]}\\[4pt]
&\implies\, g(x) = x-1\\[12pt]
&\text{Next suppose $0 \le x \le 1$.}\\[6pt]
&\text{Then by previous logic}\\[4pt]
&\phantom{\implies\,}
g(x)
=
\begin{cases}
1-x &\text{if}\;\;1-x \ge x^2/4\\[3pt]
x^2/4 &\text{if}\;\;x^2/4 > 1-x
\end{cases}
\\[6pt]
&\text{Equivalently, for $0 \le x \le 1$}\\[4pt]
&\phantom{\implies\,}
g(x)
=
\begin{cases}
1-x &\text{if}\;\;0 \le x \le 2\sqrt{2}-2\\[3pt]
x^2/4 &\text{if}\;\;2\sqrt{2}-2 < x \le 1
\end{cases}
\\[12pt]
&\text{Collecting the results obtained so far}\\[6pt]
&\phantom{\implies\,}
g(x)
=
\begin{cases}
1-x &\text{if}\;\;x \le 2\sqrt{2}-2\\[3pt]
x^2/4 &\text{if}\;\;2\sqrt{2}-2 < x \le 2\\[3pt]
x-1 &\text{if}\;\;x >2
\end{cases}
\\[12pt]
&\text{Looking at each of the $3$ pieces in turn},\\[2pt]
&\text{it follows that $g$ is}\\[2pt]
&\phantom{\implies\,}{\small{\bullet}}\;\;\text{decreasing on the interval $(-\infty,2\sqrt{2}-2]$},\\[2pt]
&\phantom{\implies\,}{\small{\bullet}}\;\;\text{increasing on the interval $(2\sqrt{2}-2,\infty)$}\\[4pt]
&\text{hence the minimum value of $g$ is}\\[4pt] 
&\phantom{\implies\,}\;\,g(2\sqrt{2}-2)\\[4pt]
&\phantom{\implies\,}=1 - \left(2\sqrt{2}-2\right)\\[4pt]
&\phantom{\implies\,}=3 - 2\sqrt{2}
\end{align*}
