Which point is not a corner of the feasible region of the system of inequalities?

Question

x+2y<=10
3x+y<=10
x>=0,y>=0

a.(10/3,0)
b.(2,4)
c.(0,5)
d.(10,0)

Answer 1

u have to draw a graph to easily identify this.

or else solve the 2 equations simultaneously

multiply equation 1 with 3, it results in
3x+6y<=30
now subract this equation from the 2nd equation

3x+6y<=30
3x+y<=10
------------------
0x+5y<=20
y<=4
if y is 4 then x is
x+2(4)<=10
x+8<=10
x<=2
So (2,4) is a corner point

In eqn 1
if x 0 4
y 5 3


In eqn 2
if x 0 1
y 10 7

if u draw these two lines & shade parts below them they are the feasible region.

check out the ends

the answer for ur question is d

Answer 2

You only have a small number of choices test them all.

You will find that c: (0,5) gives a value less than 10 in the second equation. It is therefore not on a boundary.

Answer 3

d.(10,0)
It is not in the feasible region.