Offtopic-Keep yourself occupied and help me :)

Transfer Students
41

<p>Citan: Here's one: Let f(x,y) = 1/(y+x-.5). This function will be continuous over the entire xy domain, EXCEPT where y = .5 - x (i.e a line parallel to y = 1 - x). Note that this line goes through the tip of triangle but not the trapazoid. Hence, the volume of the tip is undefined and we get the wrong answer.</p>

<p>There are many others. Just find a function that fails anywhere in the tip, and you have shown that subtracting the tip from the triangle is unreliable.</p>

<p>Edit(for Citan): The reason this confusion is occuring is because you are messing with our system of writing integrals. I understand that your system of 'order of operations' is valid, but please use our convention of 'intg' or 'int' to mean the integral sign and write out the equations. It will save a lot of headaches.</p>

42

<p>Ok, to prevent all confusion,
here are the only 2 possible ways in which we can write the original problem:</p>

<p>intg[0 to 1]intg[1-y to 4-y] f(x,y) dxdy + intg[1 to 4]intg[0 to 4-y] f(x,y) dxdy
intg[0 to 1]intg[1-x to 4-x] f(x,y) dydx + intg[1 to 4]intg[0 to 4-x] f(x,y) dydx</p>

<p>Examples of incorrect notation:
intg[1-x to 4-x]intg[0 to 1] f(x,y) dxdy + intg[0 to 4-x]intg[1 to 4] f(x,y) dxdy
intg[1-y to 4-y]intg[0 to 1]f(x,y) dydx + intg[0 to 4-y]intg[1 to 4] f(x,y) dydx</p>

43

<p>ok thanks for the help everyone, i havent looked over it yet b/c i found something new but i will go over it whne i redo that problem</p>

<p>HOw do you intergrate e^x^2</p>

<p>? ???????</p>

44

<p>Here is another one, please pelase help me this is my take home final andi just want to be sure.</p>

<p>Use Lagrange multipliers to fidn the minimum value of the function f(x,y,z)=4x^2+y^2+5z^2 subject to the constraint 2x+3y+4z=12.</p>

<p>I got
x=5/11
y=30/11
z=8/11
Lamda=20/11</p>

<p>so for the min i got 120/11</p>

<p>but i think thats wrong b/c my Lambda was only positive and i only got one crit point.
can anyone help?</p>

45

<p>Malishka31: For the first question, you must use a power series expansion of e^x^2. As you know, e^x = Sigma(from 0 to infinity) (x^n)/(n!), and so e^x^2 = Sigma(from 0 to infinity) (x^2n)/(n!). Simply integrate this and you have your answer.</p>

<p>For the second, I got the same x, y, z, and lambda you got, but I think your minimum is off(check your substitution). I got 1320/121. And having a positive lambda has nothing to do with the number of critical points you may find.</p>

46

<p>where did you plug in the stuff ?</p>

47

<p>arent lagrane multipliers supposed to give two answers, one max and one min? how come i only get one answer? </p>

<p>and this is how i got 120/11 </p>

<p>f(5/11,30/11,8/11)= 4(5/11)^2+30/11 ^2 +5(8/11)^2=120/11</p>

48

<p>Q1) I don't know what you mean. The series expansion for e^x is pretty standard stuff, it's just a simple Maclaurin series. If you substitute x^2 into that equation, you end up with something you can integrate.</p>

<p>Q2) My bad. We got the same answer, yours is just simplified version of what I had.</p>

<p>And no, the method of lagrange multipliers can give you any number of solutions (since you are simply solving a system of equations).</p>

49

<p>conjunction junction whats your function</p>

<p>whereas y)x) over alpha omega x/zulu multiplicitave inverse of beta when beta is the square of the square of the apple ;)</p>

50

<p>For lagrange multipliers if you get only one solution how do you know if your getting the max or the min???????</p>

51

<p>uh...malishka...aren't you a math major?</p>

52

<p>and? </p>

<p>its math/econ</p>

<p>i am a half math major, meaning i only need to know 1/2 the info.</p>

<p>i have a 98 in calc 3, how, i dont know... i swear i dont know anything.</p>

<p>i got an 85 on this last exam and i think i made up half of my answers</p>

<p>I finally went over the first 3 pages, ... Thanks everyone who helped on that one... I split it into 6 sections, which would also be correct but oviously two would be much much more efficient.</p>

53

<p>ijflexi: lol!</p>

<p>Malishka31: To check whether it is a maxima or minima, simply dump some points into the equation that are around the critical point you got. This is dangerous to do in other contexts(ie local max/min, saddle points), but it should be ok in this question.</p>

54

<p>i plugged in f(6/11,31/11,9/11 = 12.479</p>

<p>f(4/11,29/11,7/11)= 9.504</p>

<p>our original value is 10.909</p>

<p>its right int he middle, so i dont get it? </p>

<p>I mean the graph is a paraboloid opening up toward the z axis with a plane intersecting it at an angle. </p>

<p>techinically looking at it from xy plane view there is only 1 max and one min. </p>

<p>I think i did something wrong, and i dont know what.</p>

55

<p>\frac{d}{dx}f(x) = f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{{h(x)}^2}. </p>

<p>ah ha!</p>

56

<p>yes ijflexi, its exactly like that.</p>

57

<p>i figured as much..whats really got me stumped is this :</p>

<p>/frac{s}{e}{x}y(l)=a{d}y x2 / holla</p>

58

<p>yes well if you use the integral of the quadratic forumula and then quangulate the area between the ellipsoidial parabolic hyperbola you will then achieve the desired result.</p>

59

<p>i wanna get my quangulate on wit chu then we can talk about some ellipsodial parabolic hyperbola between the sheets..</p>

<p>all in good fun</p>

60

<p>erotic text for the math major.</p>