2014-I-18

$\begin{array}{rcl}
\dfrac{y-90}{x-45} & = & \dfrac{0-90}{180-45} \\
3y – 270 & = & – 2x + 90 \\
2x + 3y & = & 360
\end{array}$

Therefore, the required system of inequalities is

$\left\{ \begin{array}{l}
6x+7y \le 900 \\
2x + 3y \le 360 \\
x \ge 0 \\
y \ge 0
\end{array}\right.$

$\left\{ \begin{array}{l}
6x+7y \le 900 \\
2x + 3y \le 360 \\
\text{$x$ and $y$ are non-negative integers}
\end{array} \right.$

Hence, the feasible solutions are the integer points of the shaded region in Figure 7. Let the total profit $\$P(x,y)=440x + 665y$. Note that the optimal solution must be one of the vertices of the shaded region in Figure 7.

Note that the $y$-intercept of $L_2$ is $120$ and the $x$-intercept of $L_1$ is $150$. Therefore, the vertices of the shaded region of Figure 7 are
$(0,0)$, $(150,0)$, $(45,90)$ and $(0,120)$.

For $(0,0)$,

$\begin{array}{rcl}
P(0,0) & = & 440(0) + 665(0) \\
& = & 0
\end{array}$

For $(150,0)$,

$\begin{array}{rcl}
P(150,0) & = & 440(150) + 665(0) \\
& = & 66\ 000
\end{array}$

For $(45,90)$,

$\begin{array}{rcl}
P(45,90) & = & 440(45) + 665(90) \\
& = & 79\ 650
\end{array}$

For $(0,120)$,

$\begin{array}{rcl}
P(0,120) & = & 440(0) + 665(120) \\
& = & 79\ 800
\end{array}$

Hence, the greatest profit is $\$79\ 800$. Therefore, the total profit cannot exceed $\$80\ 000$. I don’t agree.