Find minimum and maximum of $x^2 - 2x - y$ where $2x+3y \le 6, 2x+y \le 4; x,y \ge 0$
I am very bad with inequalities, please help
Asked
Active
Viewed 259 times
2
-
Do you allow Lagrange multiplier? – Easy Mar 14 '13 at 06:08
-
@Cameron Buie I am in 9th grade, I don't know much calculus; I don't like to use Lagrange multiplier – Xeing Mar 14 '13 at 06:10
-
1@Inceptio: Based on several of your comments, lately, you are either psychic, or possessed of some source of information to which I have no access. Kudos, either way. – Cameron Buie Mar 14 '13 at 06:13
-
@CameronBuie: Hah! Nothing like that. I was just trying to tell that why use calculus, when it is rather easy to use simple high school math? – Inceptio Mar 14 '13 at 06:18
-
@Inceptio: When you get accustomed enough to using a hammer, everything starts to look like a nail. – Cameron Buie Mar 14 '13 at 06:19
-
@CameronBuie: Well, yeah.. I can not make my point here. Anyway, peace. Lets think of the problem. – Inceptio Mar 14 '13 at 06:22
3 Answers
4
Hint: $x^2-2x-y=(x-1)^2-y-1$. Then consider the parabola $y=(x-1)^2-1$ and draw area of all the restricted conditions.
Easy
- 4,485
-
1(+1): An excellent example of where less advanced mathematical machinery can make things easier than the more advanced sort! – Cameron Buie Mar 14 '13 at 06:18
2
Below is a graphical solution to the problem. Note that $$f(x,y) = x^2-2x-y = c$$ is a parabola and our goal is to find the maximum $c$ possible.
The blue color region is the feasible region obtained using the constraints.
0
Draw the region. For the maximum, for any fixed $x$, we should pick $y$ as small as possible. This is $0$. Now we should pick $x$ as far from $1$ as possible. That's easy to pick out.
The minimum may be more complicated. The two interesting constraint lines meet at $(1.2)$. So from $x=0$ to $x=1$, the best choice of $y$ is from $2x+3y=6$. From $x=1$ to $x=3$ the best choice is from $2x+y=4$. On each interval, we then have a quadratic minimization problem.
André Nicolas
- 507,029