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Homework Help: Did I do these problems correctly?

  1. Dec 14, 2007 #1
    Problem 1:

    1. The problem statement, all variables and given/known data
    A surface consists of all points P such that the distance from P to the plane y = 1 is twice the distance from P to the point (0,-1,0). Find an equation for this surface and identify it.

    2. Relevant equations
    the distance between (0,-1,0) and any point on the surface is:

    sqrt(x^2 + (y+1)^2 + z^2)

    the formula for the distance between a point on the surface and the plane y = 1 is:

    |y-1|/sqrt(1) or just |y-1| (is this correct?)

    3. The attempt at a solution
    I came up with this for the equation of the surface:

    2*sqrt(x^2 + (y+1)^2 + z^2)=|y-1|

    squaring both sides and completing the square for the y term, I ended up with:
    3(x^2)/4 + [9(y+5/3)^2]/16 + 3(z^2)/4 = 1, which is the equation of an ellipsoid.

    Did I do this problem correctly? I wasn't 100% sure which side of the equation to put the 2 on and even if I used the right equations. I may be second guessing myself, but I would really like other peoples opinions.

    Problem 2:

    1. The problem statement, all variables and given/known data
    Find all critical points and determine whether local maximums or minimums occur at these points for the function f(x,y) = 2x^3 + xy^2 + 5x^2 + y^2

    2. Relevant equations
    first partials =
    fx(x,y) = 6x^2 + y^2 + 10x and fy(x,y) = 2xy + 2y

    3. The attempt at a solution
    I sometimes get stumped on simultaneous equations, so I wanted to see if I did this correctly. Critical points occur when both of these equations = 0 simultaneously. I got these points to be f(0,0)=local min, f(-1,-2)=saddle point, f(-1,2)=saddle point, f(-5/3,0)=local max

    Was wondering if I did this correctly?

    Problem 3:

    1. The problem statement, all variables and given/known data
    Find the absolute maximum and absolute minimum values of the function f(x,y)=x^2 - y^2 +xy - 3x on the triangle in the plane whos vertices are (0,0), (3,0), and (3,3)

    2. Relevant equations
    well, again here I got the partials to be
    fx(x,y) = 2x + y - 3 and fy(x,y) = -2y + x

    3. The attempt at a solution
    Here I was somewhat confused by the critical points obtained from the line y=x of the triangle. I ended up with 9 critical points to check for absolute max/mins. Here's what I did:

    I set the first partials equal to 0 simultaneously and got just 1 point, f(6/5, 3/5)

    I took the bottom of the triangle and noticed 0 < or = x < or = 3, and y=0. Puting y=0 into our original function gives f(x,0)=x^2 - 3x. Taking the derivative of this and setting it =0 gave me the point f(3/2,0). Then, looking at the boundaries of this line, you get the points f(0,0) and f(3,0).

    In a similar manner for the right side of the triangle, I got the three points f(3,3/2), f(3,0) (which we already have) and f(3,3). So now, I am up to 6 points.

    The final line is what was tricky for me. It's equation is y=x, x varies from 0 to 3 inclusive and y also varies from 0 to 3 inclusive. So I thought I had to check the boundary points of f(x,x) and f(y,y).

    So for f(x,x), the original function = x^2 - 3x, whose derivative is = 0 when x=3/2. this gave me the point f(3/2,3/2),(I think?). So now(here is where I am not positive in my reasoning) I plugged in the two y values on the boundary and got f(3/2, 0) (already have this) and f(3/2, 3) (which is not in domain (not inside triangle), so we throw it out). Was it a mistake to even consider this one?

    for f(y,y), it is similar. we get the original function = y^2 - 3y so y'=0 when y=3/2. this gives the critical point f(3/2,3/2), which we already have. Plugging in the x values on the boundary we get the two points f(0,3/2) (which is not in domain so we throw it out and f(3,3/2) (which we already have).

    So, I got the following critical points:

    f(3/2, 0)
    f(0,3/2), not in domain
    f(3/2,3), not in domain

    After checking these, I got f(3/2,0) and f(3/2,3/2) = -9/4 = absolute minimums and f(3,3/2) = 9/4 = the only absolute maximum. Did I do anything wrong here? I feel like I may have found an extra critical point mistakenly in finding the critical points of the third line.

    And lastly, problem 4:

    1. The problem statement, all variables and given/known data
    A box without a lid is to have a volume of 128,000cm^3. Find the dimensions that minimize the amount of cardboard used.

    2. Relevant equations
    Well, I used lagrange multipliers, which is how we were supposed to do it.

    3. The attempt at a solution
    our function we wish to minimize is:
    f(x,y,z) = xy + 2xz + 2yz

    our limitation on its domain is:
    g(x,y,z) = xyz = 128000

    Taking the first partials of each and setting the gradient of f = lamda times the gradient of g, I got the following equations:

    and xyz=128000 (our limitation on the domain)

    solving these simultaneous equations, I got x=y=2z. then, solving for one of the variables, I got x=y=cube root of 256000=2z.

    I was confident here except for this: why is this a minimum volume and not a maximum volume? did I miss a critical point in solving the simultaneous equations?

    Sorry the post is so long, and thanks to anyone for the help. I hope I asked questions adaquately and provided the necessary info.
    Last edited: Dec 14, 2007
  2. jcsd
  3. Dec 15, 2007 #2

    Gib Z

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    Homework Helper

    For problem 4, I don't think it was a minimum volume or maximum volume, its the minimum material used.
  4. Dec 15, 2007 #3
    Yes, 1 seems all right to me.
  5. Dec 15, 2007 #4
    Thanks for the comments guys. Sorry, I meant a minimum material used for #4. I'm still not sure why it's a minimum material instead of a maximum material since lagrange multipliers finds max and mins...Anyway, thanks for the replies.
  6. Dec 15, 2007 #5

    Gib Z

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    Homework Helper

    Thats exactly correct, you don't know if its a maximum or minimum. However, we do know that the function is continuous over the interval in question, which means we can apply the mean value theorem. We test another value in that interval, if its larger than our value, then our value is a minimum, if its smaller than our value, then its a maximum.
  7. Dec 16, 2007 #6
    how about 2 and 3?
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