Sushi Posted July 17, 2011 Share Posted July 17, 2011 (1) g(x,y) = x + y D={(x,y):x^2 + y^2 = 1, x>=0 , y>=0｝ (2) g(x,y) = x^2 + 2*y^2 - x D={(x,y): x^2 + y^2<=1 } i must solve local maximum and local minimum please write method of solution Quote Link to comment Share on other sites More sharing options...

Slayerz Posted July 17, 2011 Share Posted July 17, 2011 (edited) cant do it Edited July 17, 2011 by Slayerz Quote Link to comment Share on other sites More sharing options...

Jorz Posted July 17, 2011 Share Posted July 17, 2011 wtf i googled 'japanese dictionary' and this came up o0 Quote Link to comment Share on other sites More sharing options...

Tarzan Posted July 17, 2011 Share Posted July 17, 2011 looks like the people here will be of help But sorry, I don't know either Quote Link to comment Share on other sites More sharing options...

JohnWayne Posted July 17, 2011 Share Posted July 17, 2011 (edited) change the polynomials to binomials - factor out the binomials and so on hehehe I finished college about 30 years ago, you will need to be able to solve this just to get out of the parking lot. Edited July 17, 2011 by JohnWayne Quote Link to comment Share on other sites More sharing options...

Wizbang Posted July 17, 2011 Share Posted July 17, 2011 7 and 3 Quote Link to comment Share on other sites More sharing options...

amanda Posted July 17, 2011 Share Posted July 17, 2011 Aw, basic algebra! How cute. Quote Link to comment Share on other sites More sharing options...

Saizou Posted July 17, 2011 Share Posted July 17, 2011 You have to look for first and second partial derivations, first to determine points where it's 0, which are possible minimums/maximums and second to determine if it's minimum (positive) or maximum (negative). In case a first derivation by x and derivation by y are both constant (1), and second derivation is 0, so I'd go look what the domain of the function is (part of a circle where both x and y are positive), then I'd sketch what I have and look what values would be on that arc endpoints and point in the middle of that arc and get the minimum/maximum values, so local minimums would be in g(1,0)=1 and g(0,1)=1 and local maximum would be g(sqrt(2)/2,sqrt(2)/2)=sqrt(2). In case b first derivation by x is g'(x,y)=2x and second is g''(x,y)=2 so 0 is x coordinate of a local minimum. First derivation by y is g'(x,y)=4y and second is g''(x,y)=4 so 0 is y coordinate of a local minimum. For maximum I'd go again with the same way with defining local maximum, looking up the domain of the function, and I took few points on the circle g(1,0)=1, g(0,1)=2 and g(sqrt(2)/2,sqrt(2)/2)=3/2 so local maximum would be in g(0,1)=2. Note that circle this time spans through all 4 quadrants but as x and y are both squared you can check only 3 points instead of 8. Explanation is a bit messed up but it's the best I could come up at 1am writing this all out of my head. Quote Link to comment Share on other sites More sharing options...

Gubbi Posted July 18, 2011 Share Posted July 18, 2011 In case b first derivation by x is g'(x,y)=2x and second is g''(x,y)=2 so 0 is x coordinate of a local minimum. If i read the 2nd right it was g(x,y) = x^2 + 2y^2 - x. This leads to a g'(x,y) = [2x - 1, 4y] So your local extrema is at x = 1/2. The rest you can see from Saizou Quote Link to comment Share on other sites More sharing options...

rolf Posted July 18, 2011 Share Posted July 18, 2011 @Saizou: The first one is, as far as I can see it, correct. I can't really proof with the first quadrant though, but I can't seem anything wrong with it. However, it's not really a mathematical proof ---- For the second, it's a bit more complex. You assume that the maximum value is not dependent on both x and y, that is wrong. Furthermore, you have to check all 4 quadrants, the fact that the domain is squared is not sufficient, the function is different on different parts of the quadrant. Anyway, my version: g(x,y) = x^2 + 2y^2 - x g_x(x,y) = 2x - 1, g_y(x,y) = 4y Hence, we have a critical point at (1/2, 0). We continue to check what it is, by filling it in. g(1/2, 0) = -1/4, which is a minimum (but we only observe that later in the proof) Since it has a critical point in the circle, and the functions are differentiable everywhere, we can check for only the boundaries of the domain (somehow this is not exactly valid, but valid for this case). We rewrite the domain to x = cos t y = sin t with -pi < t <= pi We define a function f(t) s.t. f(t) = g(cos t, sin t) = cos^2 t + 2 sin^2 t - cos t = 1 + sin^2 t - cos t We evaluate f'(t) s.t. f'(t) = 2 sin t cos t + sin t = sin t ( 1 + 2 cos t ) The critical points are where f'(t) = 0, hence sin t = 0 OR 1 + 2 cos t = 0 t = 0 OR t = pi OR cos t = 1/2 t = 0 OR t = pi OR t = pi / 3 OR t = 2pi / 3 Simple fill in f(0) = 0 f(pi) = 0 f(pi/3) = 11/4 f(2pi/3) = 15/4 Observe that f(2pi/3) is larger than f(pi/3), but since the functions are declining in between, f(pi/3) is also a _local_ maximum. If you would like the maximums, only the f(2pi/3) would be good. Furthermore, we see that the line for f(0) and f(pi) is the equation y=0, the same over which the critical point (-1/2,0) goes. If we check this line from the given function g, we get the function g(x,0) = x^2 - x, of which we can easily see that it only has 1 minimum; the critical point at x=1/2. Hence, the f(0) and f(pi) are not local minimums, there is a declining line towards (1/2,0). Quote Link to comment Share on other sites More sharing options...

Saizou Posted July 18, 2011 Share Posted July 18, 2011 I haven't seen that -x at the end in the second case so it's just wrong from the start, my bad First one isn't meant to be a proof, just that you try see what you have and try to conclude what to look at. And in this case you are supposed to look in the first quadrant only because of x>=0, x>=0 part. I'd say the points on the arc I took could be the dubious part, that would need better explanation. Rolf's version makes far more sense than mine, substitution is the way to go Quote Link to comment Share on other sites More sharing options...

rolf Posted July 18, 2011 Share Posted July 18, 2011 No problem, we can learn from each other. The first exercise isn't indeed a proof, but I would solve it in the same way as you did. The first exercise cannot be solved by my method, since x^2 + y^2 = 1 for x>=0, y>=0 is not differentiable everywhere. So use your intuition about the Euclidean and Manhattan distance and it'll work Quote Link to comment Share on other sites More sharing options...

JohnWayne Posted July 18, 2011 Share Posted July 18, 2011 dang I wish I would have had you all to help me get through engineering school ... where was you 30 years ago ... maybe not even born yet ..hehehehe lots of smart folks here Quote Link to comment Share on other sites More sharing options...

kristy Posted July 18, 2011 Share Posted July 18, 2011 holy shit... what math is this? calc? Quote Link to comment Share on other sites More sharing options...

rolf Posted July 18, 2011 Share Posted July 18, 2011 (edited) @JohnWayne: A well, at least now you can help others. Right? And no, I wasn't born 30 years ago @kristy: Calculus yeah. Actually, the subject of Section 13.2 (Extreme Values of Functions defined on Restricted Domains) of the book Calculus by Adams. To give an impression, 13.2 only exists of 6 pages. Total number of pages is about 1000, 17 chapters and each of approximately 8 sections. Although this is interesting, it's not really my subject (hence, I guarantee nothing about correctness). But it's not that hard if you understand the basics. It took me about 15 minutes to recapture my old lectures I've long forgotten and about 5 minutes to solve it. Edited July 18, 2011 by rolf Quote Link to comment Share on other sites More sharing options...

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