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:

(10, 50)

$5 x+y ≤ 100, x+y ≤ 60$

$x ≥ 0, y ≥ 0$ → region is in first quadrant

plotting lines

$5 x+y=100$

$x+y=60$

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

$A(0,0)$

$B(0,60)$

$C(10,50)$

$D(20,0)$