# 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.

• First try to simplify the expression for $g(x)$ (i.e., avoid using $\max$, use a case destinction if necessary) – Hagen von Eitzen Apr 28 '17 at 6:02
• I am getting two functions for two intervals $y(y - x)$ and $y(x - y)$ – Anshu Singh Apr 28 '17 at 6:10

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 .

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 $g(2(\sqrt{2}-1)) = 1- 2(\sqrt{2}-1) = 3-2\sqrt{2}$.

\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*}