EduFarming
Your list · 0हिं

Type at least two letters. Press Esc to close.

↑↓ navigate↵ openTAB cycle filter

Linear Programming · Question 57 of 59

Question 57

HardOriginal question

What can be said about: maximise Z=x+yZ = x + y subject to x+y≤3x + y \le 3, 2x+3y≥122x + 3y \ge 12, x≥0x \ge 0, y≥0y \ge 0?

Options for question 57

Show the answer and explanation
CorrectNot quiteAnswer B

Correct answer: B

Explanation

  1. On x+y≤3x + y \le 3 in the first quadrant, 2x+3y2x + 3y is largest at (0,3)(0, 3), where it is 9.

  2. 9 is less than 12, so no point satisfies both constraints: the feasible region is empty.

Worth noting

(0,3)(0, 3) satisfies x+y≤3x + y \le 3 but gives 2x+3y=9<122x + 3y = 9 < 12; (0,4)(0, 4) breaks x+y≤3x + y \le 3.

Concept
An infeasible problem
Typical time
120 s
Your time
—
The rule this testsUnbounded and infeasible problems
Practise this chapter →

Ten questions from Linear Programming, at your own pace.

Practise this chapter →

Ten questions from Linear Programming, at your own pace.