Offtopic-Keep yourself occupied and help me :)

Transfer Students
61

<p>Wow, you guys are having way too much fun with math talk.</p>

<p>Anyway, Malishka31, regarding post #54, I don't know how you can classify it as a parabaloid (since the graph of the function is 4-dimensional, it is very difficult to visualize). Using level surfaces, my version of the function is 'ellipsoids within ellipsoids' that merge to create the overall graph(like I said, difficult to visualize!). This magical shape is being intersected by the three-dimensional plane you described, and that intersection is the basis of the problem. With wacky shapes like this, it is important that you resist the urge to think of max and min in the traditional sense, since we have no idea what this would look like in 4-dimensional space.</p>

<p>Moving on, the advice I gave you needed a bit more explanation (sorry, I took this course in fall '05 and I'm helping you in order to refresh my knowledge :) ). The reason we can't 'pull points out of our ass', so to speak, is because the points you listed DO NOT lie on the restriction(ie plane) that was specified in the initial problem. Try making up random points that lie on the plane, and then dump them into the equation, and you should be able to justify to yourself that it is indeed a minima.</p>

<p>Note: On a test, you would likely have to do something more rigourous than this to justify it is a minima. Your best bet would be the second derivatives test.</p>

62

<p>ok lets pretend i am reatarted.</p>

<p>how do i make up points that lie on the plane?</p>

<p>and being on the plane doesnt necessarly mean its on that shape as well.</p>

<p>and i got a paraboloid by solving for z and then graphing, (dont ask why i did that)</p>

63

<p>haha and im a business major lol</p>

64

<p>lol, Malishka31, don't be so hard on yourself. Math is not an easy subject, even for those who are smart enough to major in it.</p>

<p>Anyway, as you already probably guessed, the point we got for our answer (5/11,30/11,8/11), lies on our plane. This is because we were solving for extrema on the function f(x,y,z) constrained to the plane described in the problem. To get points close to the minima that lie on the plane, you simply need to dump two values that are close to the minima into the plane equation and solve for the third value. For example, lets say I chose x=5/11 and y=31/11 (x is the same, y is increased by 1). So we dump these two values into 2x+3y+4z=12 and find that z=29/44. Substituting this point into f(x,y,z) gives 10.32ish, which is obviously larger than 120/11. Checking other points in this manner, it is easy to convince yourself that the point we found is indeed the minima.</p>

<p>In regards to your second point, I believe you are mistaken. EVERY point on that plane WILL be a part of the function f(x,y,z). Look at the domains; while the plane only has two degrees of freedom(since selecting two arbitrary values will define the third), f(x,y,z) can take on ANY value in R^3 space. So, any arbitrary point you select on the plane will be in the domain of f(x,y,z).</p>

<p>I don't know how to answer the third part. You shouldn't try to solve for z. f(x,y,z) is a function; it has one(and only one) value for every point in R^3 space. For example, at the point (1,1,1), the function f(x,y,z)=10. The concepts of functions, domains, and graphing all carry over from previous single variable courses, we just have to be more careful when applying them to higher order spaces.</p>

<p>If, however, you have been able to graph this 4 dimensional function, please write a scientific paper on it and send it to me so I can publish it under my name..er I mean read it. <em>shifty eyes</em></p>