Suppose that an integer $x$ and a real number $y$ satisfy both the equation $$2 ( y + 1 ) = x ( 8 - x ) \tag{1}$$ and the inequality $$5 x - 4 y + 1 \leqq 0 . \tag{2}$$ We are to find $M$, the maximum value of $y$, and $m$, the minimum value of $y$. First of all, let us transform (1) into $$y = - \frac { 1 } { \mathbf { P } } ( x - \mathbf { Q } ) ^ { 2 } + \mathbf{R} .$$ Also, from (1) and (2) we obtain the inequality in $x$ $$2 x ^ { 2 } - \mathbf { S T } x + \mathbf { U V } \leqq 0 . \tag{3}$$ Thus when $x$ is an integer satisfying (3) if we consider the range of values which $y$ can take, we see that $y$ is maximized at $x = \square \mathbf{ V }$ and is minimized at $x = \square \mathbf{ W }$, and hence that $$M = \mathbf { X } , \quad m = \frac { \mathbf { Y } } { \mathbf{Z} } .$$
Suppose that an integer $x$ and a real number $y$ satisfy both the equation
$$2 ( y + 1 ) = x ( 8 - x ) \tag{1}$$
and the inequality
$$5 x - 4 y + 1 \leqq 0 . \tag{2}$$
We are to find $M$, the maximum value of $y$, and $m$, the minimum value of $y$.
First of all, let us transform (1) into
$$y = - \frac { 1 } { \mathbf { P } } ( x - \mathbf { Q } ) ^ { 2 } + \mathbf{R} .$$
Also, from (1) and (2) we obtain the inequality in $x$
$$2 x ^ { 2 } - \mathbf { S T } x + \mathbf { U V } \leqq 0 . \tag{3}$$
Thus when $x$ is an integer satisfying (3) if we consider the range of values which $y$ can take, we see that $y$ is maximized at $x = \square \mathbf{ V }$ and is minimized at $x = \square \mathbf{ W }$, and hence that
$$M = \mathbf { X } , \quad m = \frac { \mathbf { Y } } { \mathbf{Z} } .$$