2009-I-16

1. Note that $L_1$ passes through $(12,24)$ and $(8,16)$. Then the equation of $L_1$ is

   $\begin{array}{rcl}
   \dfrac{y-24}{x-12} & = & \dfrac{16-24}{8-12} \\
   \dfrac{y-24}{x-12} & = & 2 \\
   y-24 & = & 2x – 24 \\
   2x – y & = & 0
   \end{array}$

   Note that the slope of $L_1$ is $2$. Since $L_1 \perp L_2$, then the slope of $L_2$

   $\begin{array}{cl}
   = & -1 \div 2 \\
   = & \dfrac{-1}{2}
   \end{array}$

   Therefore, the equation of $L_2$ is

   $\begin{array}{rcl}
   y -24 & = & \dfrac{-1}{2} (x – 12) \\
   2y – 48 & = & -x + 12 \\
   x + 2y – 60 & = & 0
   \end{array}$

2. The required system of inequalities

   $\left\{ \begin{array}{l}
   x \ge 8 \\
   y \ge 10 \\
   2x-y \ge 0 \\
   x + 2y -60 \le 0
   \end{array} \right.$

$\left\{ \begin{array}{l}
x \ge 8 \\
y \ge 10 \\
y \le 2x \\
4x + 8y \le 240
\end{array} \right. $

After simplifying the above system, we have

$\left\{ \begin{array}{l}
x \ge 8 \\
y \ge 10 \\
2x – y \ge 0 \\
x + 2y – 60 \le 0
\end{array} \right.$

Note that the area contains all feasible solutions is the shaded region in the figure. Consider the intersection point of $L_2$ and $L_4$, sub. $y=10$ into $x+2y – 60=0$, we have

$\begin{array}{rcl}
x + 2(10) – 60 & = & 0 \\
x & = & 40
\end{array}$

Therefore, the four vertices are $(12,24)$, $(8,16)$, $(8,10)$ and $(40,10)$. Let $\$P$ be the total profit. Then we have $P(x,y)=4000x + 6000y$.

$\begin{array}{rcl}
P(12,24) & = & 4000(12) + 6000(24) \\
& = & 192~000 \\
P(8,16) & = & 4000(8) + 6000(16) \\
& = & 128~000 \\
P(8,10) & = & 4000(8) + 6000(10) \\
& = & 92~000 \\
P(40,10) & = & 4000(40) + 6000(10) \\
& = & 220~000
\end{array}$

Therefore, the greatest profit according to the constraints is $\$220~000$. The claim is not agreed.