Question

Pre-calculus question

Answer 1

@hwyl

Answer 2

Hmm... I would say, The feasible set is unbounded (and closed,

because the constraints are not strict inequalities). \(3x+y<= 5\) implies \(x <= (5 - y)/3\). Let, \((y-> 00\) and putting \(x = (5 - y)/3\), we satisfy all constraints , while \(x -> -00 \) and \(y -> 00.\) @vera_ewing

Answer 3

So what would the answer be?

Answer 4

. unbounded

Answer 5

lol, you already answered it in chat... do u forget things fast ?