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Minimizing the area in a parabola

  1. Oct 21, 2005 #1
    Let f(x) = 6-x^2 for 0<w<sqrroot(6); let A(w) be the area of the triangle formed by the coordinate axes and the line tangent to the graph of f.
    see figure : http://img472.imageshack.us/img472/50/a5nt.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 21, 2005 #2


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    Okay, I looked at your picture and read your post. There is no question there! I can guess from the title that the problem is to find the point on the parabola such that the triangle constructed has minimum possible are- though that is NOT "minimizing the area in a parabola"!

    Have you done anything at all on this? In particular, taking the "x" coordinate of the point on the parabola as fixed (call it x0, say) can you write the equation of the tangent line there?
    Once you have that, you should be able to write the equation for the area of the triangle as a function of x0.
  4. Oct 21, 2005 #3
    yea i did try it like 20x but couldnt figure it out..
    i derived to get the tangent line, so slope = -2x at point x..
    so i find the equation for the line.. thus finding b as 6+x^2
    i got this by plugging (x,6-x^2) into the equation y = -2x(x) + b and solved for b.
    then i get the final equation of y = 6-x^2! i end up where i started from! what is going on?!
  5. Oct 21, 2005 #4


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    The equation for a tangent line to a curve f(x) at a point (a,b) is given by y-b = f'(a)(x-a). Note that with f'(a), I mean the derivative of f(x) evaluated in the point (a,b) so with x = a.
    So instead of (a,b), you know have the general point (p,6-p²). You already had the derivative, so can you know get to the tangent line for a general point on the parabola? It's just filling in.

    After that, how would you find the area of the triangle? You're "lucky" since it's a right triangle ;)
  6. Oct 21, 2005 #5
    if i just find the tangent line equation, i end up with y=6-x^2 as my line of my tangent line. Is that suppose to be right?
    and how am i suppose to find the minimum area under the triangle and prove that its the smallest possible area of the triangle on the parabola function thats given?
  7. Oct 21, 2005 #6


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    In the equation I gave you, don't fill in (x,6-x²) since x is also your unknown in the equation, your variable! Instead, use (e.g.) (p,6-p²). Just apply the formula I gave. Then, for every p, you get a point (p,6-p²) on the parabola with the tangent line you just set up, p being substituted with the value in question.
  8. Oct 21, 2005 #7
    so.. y-(6-p^2) = f'(p) * (x - p)
    y - 6 + p^2 = -2p (x-p)
    y = p^2 -2px + 6 ? what does this mean? the equation of the tangent line?
  9. Oct 21, 2005 #8


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    Correct, for every p you fill in, you'll now get the tangent line to the parabola at the point x = p, y = 6-p^2.
  10. Oct 21, 2005 #9
    ok, so.. what i did was.. set 0=-2px + p^2 + 6, then solved for x. x= (p^2+6)/2p;
    where as the y-intercept is (0, p^2 +6)
    so to get the area.. A(p) = 1/2 * (p^2+6)/2p * (p^2+6)
    A(p) = (p^4 + 12p^2 + 36) / 4p
    I derive this area to find the min/max..
    am i on the right track?
  11. Oct 21, 2005 #10


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    Yes, that looks good.

    One advice though. It's generally easier (in my opinion) to not expand the brackets. So just keep A(p) look like:
    Taking the derivative will be easier in this form
  12. Oct 21, 2005 #11
    thanks alot.. i got A'(w) = 0 where w=-6 and 2.
    thus A(2) = 12.5; which is the min point im guessing.
    i can check the value of A''(2) to see if its <0 or >0 to det. if its min or max but im too lazy. i think i did this right right now. once again, thank you all for the help.
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