How many corner points does the feasible region of x + 2y ≤ 12, x ≤ 6, y ≤ 5, x ≥ 0, y ≥ 0 have?

Linear Programming · Question 27 of 59

Question 27

MediumOriginal question

How many corner points does the feasible region of $x + 2y \le 12$ , $x \le 6$ , $y \le 5$ , $x \ge 0$ , $y \ge 0$ have?

3

Correct answerYour answer

4

Correct answerYour answer

5

Correct answerYour answer

6

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

The corners are $(0, 0)$ , $(6, 0)$ , $(6, 3)$ , $(2, 5)$ , $(0, 5)$ .
The line $x + 2y = 12$ cuts off the corner $(6, 5)$ of the box, leaving 5 corners.

Worth noting

4 counts the box $x \le 6$ , $y \le 5$ alone; $(6, 5)$ is not feasible, since $6 + 10 > 12$ .

Ten questions from Linear Programming, at your own pace.

Ten questions from Linear Programming, at your own pace.