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Math Help - 3 questions

  1. #1
    Newbie Nick744's Avatar
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    3 questions

    Hey guys,

    Got some challange questions in my book, and they're challanging (for me atleast).

    1. Sketch the curve y = x(x-2)^3 , showing any stationary points and inflexions.

    2. Find the greatest and least values of the function f(x) = 4x^3 - 3x^2 - 18x in the domain -2 <= x <= 3.

    3. Two circles have radii r and s such that r + s = 25. Show that the sum of areas of the circles is least when r = s.

    Thanks very much in advance.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Nick744 View Post
    Hey guys,

    Got some challange questions in my book, and they're challanging (for me atleast).

    1. Sketch the curve y = x(x-2)^3 , showing any stationary points and inflexions.
    first this has zeros a x=0, and x=2, the latter of multiplicity 3, so we know
    that the curve looks like (x-2)^3 close tp x=2, and so we expect this point
    to be a point of inflection.

    Also as x goes to +/- infinity y goes to +infinity.

    Now lets look for the stationary points:

    dy/dx= (x-2)^3 + 3 x (x-2)^2 = (x-2)^2 [(x-2) + 3x]=2 (x-1) (2x -1)

    Which is equal to zero when x=1/2, and when x=2. Then we can deduce
    from shape of the cureve and the position of the zeros of y, that x=1/2 is a
    minimum, and x=2 a point of inflection (we could confirm this with the
    second derivative test if we wanted).

    y~=-1.68, at the minima corresponding to x=1/2.

    Putting this all together we get something like the attachment.

    RonL
    Attached Thumbnails Attached Thumbnails 3 questions-gash.jpg  
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Nick744 View Post
    2. Find the greatest and least values of the function f(x) = 4x^3 - 3x^2 - 18x in the domain -2 <= x <= 3.
    A function differentiable function f(x) has a global maximum (minimum) on a closed interval
    either at an end point of the interval, or as a calculus type local maximum in the interval.

    So here

    1. Compute the f(-2) and f(3)

    2. Differentiate f(x) to get f'(x), solve f'(x)=0 to find the stationary points. Let the stationary points in [-2, 3] be x1, .. xn.

    3. Compute f(x1), .. , f(xn), then the global maximum is:

    max[ f(-2), f(3), f(x1), .. , f(xn) ].

    The same procedure also finds the global minimum if max is rellaced by min in the last line above.

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Nick744 View Post
    3. Two circles have radii r and s such that r + s = 25. Show that the sum of areas of the circles is least when r = s.
    Assume the circles do not overlap, then the sum of the areas is:

    A=pi r^2 + pi s^2,

    ans as r + s =25, we may substitute s = 25-r for s in the equation for A:

    A = pi r^2 + pi (25-r)^2 = pi [ 2 r^2 -50 r +225 ]

    To find the stationary points of A we differentiate wrt r to get A'(r), set this to zero to find the critical points:

    A'(r) = pi [4 r - 50 ]

    So A'(r)=0, has solution r=25/2. To show that this is a minimum we differentiate again to get A''(r)= 4 pi, which is >0, so r=25/2 is a minimum.

    As r+s=25, when r=25/2 s=25/2, so we have a minimum area sum when r=s.

    RonL
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