A maximizing linear programming problem has two constraints: 2X + 4Y < 100 and 3X + 10Y < 210, in addition to constraints stating that both X and Y must be nonnegative

• (0, 0), (50, 0), (0, 21), and (20, 15)

• We first need to make a graphical representation of the constraints by computing the coordinates of the constraints.

2X + 4Y ≤ 100

            @ X=0

                        2X + 4Y ≤ 100

                        2(0) + 4Y = 100

                        Y = 100/4

Y = 25

            @ Y=0

                        2X + 4Y ≤ 100

2X + 4(0) = 100

X = 100/2

X = 50

The coordinate for 2X + 4Y ≤ 100 is (50,25).

3X + 10Y ≤ 210

            @ X=0

                        3X + 10Y ≤ 210

                        3(0) + 10Y = 210

Y = 210/10

Y = 21

            @ Y=0

                        3X + 10Y ≤ 210

3X + 10(0) = 210

X = 210/3

X = 70

The coordinate for 3X + 10Y ≤ 210 is (70,21).

• We can now identify the corner points or the extreme points of the feasible region. Before doing so, we first need to determine the feasible region. The shaded region in the graph below is the feasible region because we can see that the signs of the constraints are ≤, and having this sign means that the feasible region is towards the left of the constraint line.
• Therefore, the corner points of the feasible region are (0, 0), (50, 0), (0, 21), and (20, 15).

Please see the attached file for complete solution.