The constraints of a problem are graphed below. what are the vertices of the feasible region?

(0, 0), (0, 15), (20, 25), (40, 0)

(0, 0), (0, 40), (25, 20), (15, 0)

(0, 15), (20, 25), (40, 0)

(0, 40), (25, 20), (15, 0)

(0, 0), (0, 15), (20, 25), (40, 0)

Step-by-step explanation:

The diagram shows shaded region in the form of quadrilateral bounded with two given lines and x- axis and y-axis. Four vertices of this quadrilateral are point of intersection of those lines or lines with coordinate axes.

The first vertex of this quadrilateral lies at the origin (point of intersection of two axes), then its coordinates are (0,0).

The second vertex lies on the x-axis and its coordinates are (40,0) (placed at 40 on the x-axis)

The third vertex lies on the y-axis and its coordinates are (0,15) (placed half way between 10 and 20 on the y-axis).

The fourth vertex is the intersection of two lines. Its coordinates are (20,25) (draw vwrtical line from this vertex t x-axis, it will intersect x-axis at 20, draw horizontal line to y-axis, it will intersect the y-axis at 25).

So, correct option is A: (0, 0), (0, 15), (20, 25), (40, 0)

(0, 0), (0, 15), (20, 25), (40, 0)

Step-by-step explanation:

The feasible region is the shaded area in the graph. This region has four vertices which are:

(0, 0) at the bottom left

(0, 15) at the top left

(20, 25) at the top center

(40, 0) at the bottom right

Note: the points have the notation (x, y)

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

squares, rhombus, rectangles