If $5 x+y \leq 100, x+y \leq 60, x \geq 0, y \geq 0$. Then one of the corner points of the feasible region is:
Answer & explanation
Correct answer: option 3
$5 x+y ≤ 100, x+y ≤ 60$
$x ≥ 0, y ≥ 0$ → region is in first quadrant
plotting lines
$5 x+y=100$
|
x |
20 |
0 |
|
y |
0 |
100 |
$x+y=60$
|
x |
0 |
60 |
|
y |
60 |
0 |
checking for inequality.
checking at paint (0, 0)
for $5 x+y \leq 100 \rightarrow 0 \leq 100$ True
(solution lies to side containing $(0,0)$ )
for $x+y \leq 60 \rightarrow 0 \leq 60$ True
(Solution lies to side containing (0,0))
corner points obtained
$A(0,0)$
$B(0,60)$
$C(10,50)$
$D(20,0)$