The feasible region of L. P. P. Maximize $\mathrm{z}=70 x+50 \mathrm{y}$ subject to $8 x+5 y \leq 60,4 x+5 y…
The feasible region of L. P. P.
Maximize $\mathrm{z}=70 x+50 \mathrm{y}$ subject to $8 x+5 y \leq 60,4 x+5 y \leq 40$ and $x \geq 0, y \geq 0$
is
a triangle
a square
a pentagon
a quadrilateral
Solution
\begin{array}{|l|l|l|}
\hline Line & Point on X-axis & Point on Y-axis \\
\hline 8 x+5 y=60 & \mathrm{~A}(7.5,0) & \mathrm{B}(0,12) \\
\hline 4 x+5 y=40 & \mathrm{C}(10,0) & \mathrm{D}(0,8) \\
\hline
\end{array}
Feasible region is shaded.
Point of intersection $\mathrm{E} \equiv(5,4)$
Also $A=(7.5,0)$ and $D \equiv(0,8)$
So OAED is a quadrilateral.
