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Homework Help: Calc 1 Optimization

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider the part of the parabola y=1-x^2 from x=-1 to x=1. This curve fits snugly inside an isosceles triangle with base on the x-axis and one vertex on the y-axis. What is the smallest possible area of such a triangle?


    3. The attempt at a solution

    [tex] A = \frac{1}{2}x y[/tex]

    [tex] y = 1-x^2 [/tex]

    so

    [tex] A = \frac{1}{2}x(1-x^2)[/tex]

    I solved that, took the derivative and set it equal to 0. That gave me an answer of x=sqrt(1/3), but I know that is wrong since it doesn't encompass the parabola curve. What is wrong with this?
     
  2. jcsd
  3. Nov 24, 2008 #2
    y should be the point of the vertex not the equation of the parabola, try finding the gradient of the slopes at x=-1 and 1 and see where they cross.
     
  4. Nov 24, 2008 #3
    Of which slopes, the parabolas?
     
  5. Nov 24, 2008 #4

    Dick

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    Suppose the point of tangency is at x=a. Then you know the tangent line goes through the point (a,1-a^2) and has slope -2a, right? Work back from that and figure out where that tangent line intersects the x and y axes. Then find the area of the triangle as a function of a. Now minimize with respect to a. That's what Firepanda meant.
     
  6. Nov 24, 2008 #5
    Ok here we go:

    the slope is equal to rise over run so:
    [tex]f'(a) = \frac{y-b}{x-a}[/tex]
    where a and b are just arbitrary points.

    [tex]f'(p) = -2p = \frac{y - (1-p^2)}{x-p}[/tex]

    where p is some point. And f'(p) is the derivative of f at point p. We don't use x since x is already a unknown variable in our equation.

    If we solve that for y we get:

    [tex] y = -2p(x-p) + 1-p^2[/tex]

    [tex] y = p^2 -2px +1[/tex]

    This is our equation of some tangent line to the parabola at some point p.

    Solve for the y intercept by setting y=0
    [tex]0 = p^2 -2px +1[/tex]

    [tex]x = \frac{p^2 +1}{2p}[/tex]

    and we know the x intercept equals:

    [tex] y = p^2 + 1 [/tex]

    So back to our Area formula:

    [tex]A(p) = \frac{1}{2}(\frac{p^2+1}{2p})(p^2+1)[/tex]

    Take the derivative of that and set it equal to 0.

    Solve for p and you will get an answer of

    [tex]p=\sqrt{\frac{1}{3}}[/tex]

    Plug that back into your x and y intercept equations to get the x and y values. Plug those x and y values back into the area equation to get an answer of about 1.5396.
     
  7. Nov 24, 2008 #6

    Dick

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    Bravo! Well done! Except that I think you are forgetting the part of the triangle in the negative x region. Isn't the area of the triangle really x*y, not (1/2)*x*y? I.e. (1/2)*(2x)*y? Actually, I think you are right numerically. It's just that A(p) isn't what you said it is. Just a typo, I'm shure.
     
    Last edited: Nov 24, 2008
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