Maximization

Forum for the GRE subject test in mathematics.
spice girl
Posts: 1
Joined: Sat Dec 11, 2010 12:25 pm

Maximization

Postby spice girl » Sat Dec 11, 2010 12:28 pm

Need a little help please!!!!
Maximize the function f(x,y) =sqrt(x^2+y^2) subject to the constraints x+3y≤10,x≥0 and y≥0.

trevaskis
Posts: 22
Joined: Sun Nov 07, 2010 1:33 pm

Re: Maximization

Postby trevaskis » Sat Dec 11, 2010 12:34 pm

Look up the lagrangian multiplier method.

logaritym
Posts: 42
Joined: Sun Nov 14, 2010 10:06 am

Re: Maximization

Postby logaritym » Sat Dec 11, 2010 12:41 pm

You can just notice that the maximum is achieved when x+3y=10. Then just substitute x=10-3y in x^2+y^2 and maximize it in the interval y \in [0, 10/3]. Take a square root of the result and that's it.

brain
Posts: 28
Joined: Tue Aug 25, 2009 12:16 pm

Re: Maximization

Postby brain » Sat Dec 11, 2010 12:48 pm

First sketch the domain. It is a triangle in quarant I. Then you substitude the vertices into the given function. No need to try (0, 0) 'cause the function is bigger when x and y are getting bigger.

brain
Posts: 28
Joined: Tue Aug 25, 2009 12:16 pm

Re: Maximization

Postby brain » Sat Dec 11, 2010 3:43 pm

Spice girl, do you get f(10, 0) = 10 as an answer?

Flow
Posts: 4
Joined: Tue Dec 14, 2010 2:28 am

Re: Maximization

Postby Flow » Tue Dec 14, 2010 2:31 am

The function is simply the distance from the origin. Our domain is a triangle in the first quadrant, with the furthest vertex at (10,0). Thus, the answer is 10.




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