The constraints of a problem are listed below. What are the vertices of the feasible region? X+y<_5
y<_3
X>_0
y>_0

(0, 0), (0, 3), (2, 3), and (5, 0)

Step-by-step explanation:

The constraints of the problem are:

x + y ≤ 5

y ≤ 3

x ≥ 0

y ≥ 0

Then:

Using the second and fourth inequalities we can write:

0 ≤ y ≤ 3

Knowing that the minimum value of y is 0, then if we replace that in the first equation we get:

x + 0 ≤ 5

x ≤ 5

While for the maximum value of y, y = 3, this inequality becomes:

x + 3  ≤ 5

x ≤ 2

Now, the vertexes are the points where both variables take one of their extremes.

Then, we have a vertex at (0, 0) because we have:

x ≥ 0

y ≥ 0

So this is the vertex when both x and y take the minimum value.

When y takes the maximum value y = 3, and x takes the minimum value x = 0, we have the vertex:

(0, 3)

When y takes the maximum value, y = 3, and x takes the maximum value, x = 2, we have the vertex:

(2, 3)

When y takes the minimum value, y = 0, and x takes the maximum value, x = 5, we have the vertex:

(5, 0)

Then the four vertexes are:

(0, 0), (0, 3), (2, 3), and (5, 0)

35x 3= 105

step-by-step explanation:

35+35= 70   70 + 35= 105