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  1. #1
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    Calc help

    I need help with these fast, it's for homework and I don't get it and my homework is due at midnight tonight.

    1. Find the dimensions of the right circular cylinder with greatest volume that can be contained in a sphere of radius 1.

    2. What is the largest (by area) rectangle with sides parallel to the coordinate axes that can be contained in the region of the plane bounded by the graphs of y=0, y=x (1-E^(x/x-1)) and x=1?


    Thanks
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  2. #2
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    Quote Originally Posted by swimmerxc View Post
    ...
    1. Find the dimensions of the right circular cylinder with greatest volume that can be contained in a sphere of radius 1.

    ...
    Hello,

    1. make a sketch: see attachment

    2. Volume of the cylinder: V_{cyl}=\pi r^2 \cdot h

    3. Relation between the height h, the radius r of the base of the cylinder and the radius R of the sphere:
    R^2=r^2+\frac14 \cdot h^2 . Since R = 1 you'll get r^2 = 1-\frac14 \cdot h^2

    Plug in this term into the equation to calculate the volume of the cylinder:

    V_{cyl}(h)=\pi \left( 1-\frac14 \cdot h^2 \right) \cdot h=\pi h - \frac14 \pi h^3

    You'll get the extreme (minimum or maximum) value of V_{cyl}(h) if V_{cyl}'(h) = 0

    V_{cyl}'(h)=\pi - \frac34 \pi h^2

    \pi - \frac34 \pi h^2 = 0~\implies~h=\frac23 \cdot \sqrt{3}

    Plug in this value into the equation to calculate rē. You'll get:

    r^2=1-\frac14 \cdot \left( \frac23 \cdot \sqrt{3} \right)^2 ~\implies~r= \frac13 \cdot \sqrt{6}

    EDIT: Forgot to attach the sketch. Sorry!
    Attached Thumbnails Attached Thumbnails Calc help-zyl_inkugl.gif  
    Last edited by earboth; November 2nd 2007 at 12:22 PM.
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  3. #3
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    Quote Originally Posted by earboth View Post
    Hello,

    1. make a sketch: see attachment

    2. Volume of the cylinder: V_{cyl}=\pi r^2 \cdot h

    3. Relation between the height h, the radius r of the base of the cylinder and the radius R of the sphere:
    R^2=r^2+\frac14 \cdot h^2 . Since R = 1 you'll get r^2 = 1-\frac14 \cdot h^2

    Plug in this term into the equation to calculate the volume of the cylinder:

    V_{cyl}(h)=\pi \left( 1-\frac14 \cdot h^2 \right) \cdot h=\pi h - \frac14 \pi h^3

    You'll get the extreme (minimum or maximum) value of V_{cyl}(h) if V_{cyl}'(h) = 0

    V_{cyl}'(h)=\pi - \frac34 \pi h^2

    \pi - \frac34 \pi h^2 = 0~\implies~h=\frac23 \cdot \sqrt{3}

    Plug in this value into the equation to calculate rē. You'll get:

    r^2=1-\frac14 \cdot \left( \frac23 \cdot \sqrt{3} \right)^2 ~\implies~r= \frac13 \cdot \sqrt{6}
    thanks, number one i figured out after looking at it, however numver 2 i still dont get...

    this is how i solved #1, is it right?
    Volume=pi r^2*2h
    h=sqrt(1-r^2)

    V=2pi r^2 sqrt(1-r^2)
    V'=0
    Hence
    r^2+r-1=0
    r=(-1+sqrt(5))/2

    Height of the cylinder=2h
    =2(sqrt(1-((-1+sqrt(5))/2)^2))
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  4. #4
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    1. Find the dimensions of the right circular cylinder with greatest volume that can be contained in a sphere of radius 1.

    Draw the figure.
    It is a rectangle inscribed in the circle. Draw one radius R to one of the 4 corners of the rectangle.

    Let x = half of the base or width of the rectangle. This the radius of the cylinder.
    And h = height of the rectangle.

    By Pythagorean theorem,
    R^2 = x^2 +(h/2)^2
    So,
    (h/2) = sqrt(R^2 -x^2)

    We are looking for greatest volume so
    V = pi(x^2)h
    V = pi(x^2)[2sqrt(R^2 -x^2)]
    Differentiate both sides with respect to x,
    [R is a constant, remember],
    dV/dr = 2pi{(x^2)[-x / sqrt(R^2 -x^2)] +[sqrt(R^2 -x^2)](2x)]
    Set that to zero,
    0 = (-x^3)/sqrt(R^2 -x^2) + 2x[sqrt(R^2 -x^2)]
    0 = (-x^2)/sqrt(R^2 -x^2) + 2[sqrt(R^2 -x^2)]
    Multiply both sides by sqrt(R^2 -x^2)
    0 = -x^2 +2(R^2 -x^2)
    0 = -3x^2 +2R^2
    x^2 = 2R^2 / 3
    x = R*sqrt(2/3) ---------***

    Since R = 1,
    x = sqrt(2/3) = (1/3)sqrt(6) = 0.8165 unit

    Therefore, the greatest volume is
    V = pi(x^2)[2sqrt(R^2 -x^2)]
    V = pi(0.8165^2)[2sqrt(1^2 -0.8165^2]
    V = 2.4184 cu.units ------------------------answer.
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  5. #5
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    2. What is the largest (by area) rectangle with sides parallel to the coordinate axes that can be contained in the region of the plane bounded by the graphs of y=0, y=x (1-E^(x/x-1)) and x=1?

    y = x[1 -e^(x/(x-1))] ------is that right?

    Let me assume it is.
    So at x=1, y is not defined because of the e^(1/0). Meaning, x=1 is a vertical asymptote of y.
    [I don't know how to use any graphing calculator or computer application, so I cannot "see" the graph of y.]. But at x=0, y=0 too.

    So the rectangle has:
    height = y = x[1 -e^(x/(x-1))]
    width = (1 -x)
    Area, A = (1-x)*x[1 -e^(x/(x-1))]
    A = x[1 -e^(x/(x-1))] -(x^2)[1 -e^(x/(x-1))]
    dA/dx = .....Man!

    Continue that. Then set it to zero. Etc.
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  6. #6
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    Quote Originally Posted by swimmerxc View Post

    2. What is the largest (by area) rectangle with sides parallel to the coordinate axes that can be contained in the region of the plane bounded by the graphs of y=0, y=x (1-E^(x/x-1)) and x=1?
    Hello,

    I assume that the function reads

    f(x) = x \left(1 - e^{\frac{x}{x-1}} \right)

    According to the shape of the graph I put the width of the rectangle on the line x = 1 (see attachment)

    Let the length of the rectangle be l = 1-a. Then the area is:

    r= l \cdot w~\implies~r(a)=(1-a) \cdot f(a)

    r(a) = (1-a) \cdot a \left(1 - e^{\frac{a}{a-1}} \right) You'll get the extreme value if r'(a) = 0

    r'(a)=e^{\frac{a}{a-1}} \cdot (1-2a)+(a-a^2) \cdot e^{\frac{a}{a-1}} \cdot \frac{-1}{(a-1)^2}

    r'(a)=e^{\frac{a}{a-1}} \cdot \left(1-2a+\frac{-a(a-1)}{-(a-1)^2}\right) . Solve r'(a) = 0 for a:. Because e^{\frac{a}{a-1}} \ne 0 you have to solve:

    \left(1-2a+\frac{-a(a-1)}{-(a-1)^2}\right)=0 ~\iff~ 1-2a+\frac{a}{(a-1)}=0

    a-1 - 2a^2 +2a + a = 0~\iff~2a^2 + 4a - 1=0. I've got:

    a=-1\pm \sqrt{\frac32} . Since 0 < a < 1 only the positive result is valid.

    Thus a = -1+ \sqrt{\frac32} \approx 0.2247

    Remark: This value seems to be too small. So better check my calculations!
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  7. #7
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    Quote Originally Posted by swimmerxc View Post
    ...
    this is how i solved #1, is it right?
    Volume=pi r^2*2h
    h=sqrt(1-r^2)

    V=2pi r^2 sqrt(1-r^2)
    V'=0
    Hence
    r^2+r-1=0
    r=(-1+sqrt(5))/2

    Height of the cylinder=2h
    =2(sqrt(1-((-1+sqrt(5))/2)^2))
    Hello,

    your function is:

    V=2 \pi r^2 \cdot \sqrt{1-r^2} Now use product rule, chain rule:

    V'=\sqrt{1-r^2} \cdot 4 \pi r + 2 \pi r^2 \cdot \frac12 \cdot (1-r^2)^{-\frac12} \cdot (-2r). If V' = 0 you can multiply both sides by the square root and you'll get the equation:

    4\pi r - 4 \pi r^3 - 2 \pi r^3 = 0 which will give the result I already wrote in my previous post.
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  8. #8
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    Quote Originally Posted by earboth View Post

    Thus a = -1+ \sqrt{\frac32} \approx 0.2247

    Remark: This value seems to be too small. So better check my calculations!
    it seems to be ok, I'm going it on a math program and it runs through fine, also for 0.2247 is that the final answer or do I have to plug it back into r=(1-a)(f(a))?

    Thanks again man
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