Question

Consider the following linear programming problem:

Maximize	$12X+10Y$
Subject to:	$4X+3Y\le 480$
	$2X+3Y\le 360$
	all variables $\ge 0$

Which of the following points $(X,Y)$ could be a feasible corner point?

• A. $(40,48)$
• B. $(120,0)$
• C. $(180,120)$
• D. $(30,36)$
• E. None of these

Answer 1

The correct option is B $(120,0)$

Given constraints, $4x+3y\le 480$ and $2x+3y\le 360$

first, draw the graph for equations $4x+3y=480$ and $2x+3y=360$

for $4x+3y=480$

substitute y=0 we get, $4x=480⟹x=120$

substitute x=0 we get, $3y=480⟹y=160$

therefore, $4x+3y=480$ line passes through (120,0) and (0,160) as shown in fig.

Hence, $4x+3y\le 480$ includes the region below the line.

for $2x+3y=360$

substitute y=0 we get, $2x=360⟹x=180$

substitute x=0 we get, $3y=360⟹y=120$

therefore, $4x+3y=480$ line passes through $(180,0)$ and $(0,120)$ as shown in fig.

Hence,$2x+3y\le 360$includes the region below the line.

the shaded region as shown in figure is intersection region. From the figure, $(0,120)$ and $(120,0)$ are feasible corner points.