At which point does Z = x + 3y reach its maximum, subject to 2x + y ≤ 12, x + 2y ≤ 12, x ≥ 0, y ≥ 0?

Linear Programming · Question 40 of 59

Question 40

MediumOriginal question

At which point does $Z = x + 3y$ reach its maximum, subject to $2x + y \le 12, x + 2y \le 12, x \ge 0, y \ge 0$ ?

(4, 4)

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(0, 6)

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(6, 0)

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(0, 12)

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CorrectNot quiteAnswer B

Correct answer: B

Corners: $(0, 0)$ , $(0, 6)$ , $(4, 4)$ , $(6, 0)$ ; $Z$ there is $0$ , $18$ , $16$ , $6$ .
The largest, 18, is at $(0, 6)$ .

Worth noting

$(4, 4)$ is where the two lines cross, but $Z$ weights $y$ three times as much as $x$ , so the top corner wins. $(0, 12)$ is not feasible.

Ten questions from Linear Programming, at your own pace.

Ten questions from Linear Programming, at your own pace.