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Math Help - very easy 3 probs ...

  1. #1
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    very easy 3 probs ...

    as i am new to calculus & probability,pls show how to do these problems(may be very trivial)

    problem 1:show that the height of a right circular cylinder of the greatest volume that can be inscribed in a right circular cone is one-third of that of the cone.
    (using derivatives and limits....)

    problem 2:using integration,find the area enclosed by the curves  y= x^{2} and  y=x+2

    problem 3: out of 21 tickets marked with numbers from 1 to 21, three are drawn at random.Find the probability that the numbers on them are in A.P.



    please help!!!thanks in advance
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    2
    \displaystyle \int_{intersection}^{intersection}[f(x)-g(x)]dx
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    Quote Originally Posted by earthboy View Post

    problem 3: out of 21 tickets marked with numbers from 1 to 21, three are drawn at random.Find the probability that the numbers on them are in A.P.
    The tickets have numbers so why are we looking for letters?
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    Senior Member BAdhi's Avatar
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    1) i think my diagram is correct. if so, take a relationship between h and r from H and R which are constants

    then calculate the volume V of the cylinder and differentiate it with respect to either h (or r). Find for which value for h (or r) you get the \frac{dV}{dh}=0

    take \frac{d^2V}{dh^2}. If for each calculated value for h, you'll get a minus for \frac{d^2V}{dh^2} it means V is maximum for that value for that h
    Attached Thumbnails Attached Thumbnails very easy 3 probs ...-untitled.jpg  
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  5. #5
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    could someone write problem 2 in a little more detail(i couldnt understand)
    Quote Originally Posted by dwsmith
    The tickets have numbers so why are we looking for letters?
    sorry,but i dont get you
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    Quote Originally Posted by dwsmith View Post
    The tickets have numbers so why are we looking for letters?
    A.P. <=> Arithmetic Progression

    CB
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    Senior Member BAdhi's Avatar
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    substitute g(x)=0, f(x)=x^2 tha'll give you the area under the curve
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  8. #8
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    Quote Originally Posted by earthboy View Post
    as i am new to calculus & probability,pls show how to do these problems(may be very trivial)

    problem 1:show that the height of a right circular cylinder of the greatest volume that can be inscribed in a right circular cone is one-third of that of the cone.
    (using derivatives and limits....)
    Let the height of the cone be "h" and the radius of the base "R". If you set up a coordinate system so that the origin is at the center of the base and the axis is along the positive z-axis, then the slant side, in the xz-plane, passes through (0, 0, h) and (R, 0, 0) and so has equation z= h(1- x/R), y= 0.

    A cylinder of radius r, with center at (0, 0, 0) and axis along the positive z-axis, inscribed in that cone will go up to that line at x= r; its height is h(1- r/R) and so its volume is [itex]\pi r^2(h(1- r/R))= \pi h (r^2- r^3/R)[/itex]. That will have maximum volume when its derivative is 0- [itex]\pi r^2h (2r- 3r^2/R)= 0[/itex]. [itex]r(2- 3r/)= 0[/itex] when r= 0 (the minimum volume) or r= (2/3)R.

    Now, the height is given by h(1- r/R)= h(1- 2/3)= (1/3)h.
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    Hello, earthboy!

    Problem 3 is intriguing.


    Problem 3: Out of 21 tickets marked with numbers from 1 to 21, tjree are
    drawn at random. Find the probability that the numbers on them are in A.P.
    There are: . \displaystyle _{21}C_3 \:=\:{21\choose3} \:=\:\frac{21!}{3!\,18!} \:=\:1330\text{ possible outcomes.}


    How many of them are in Arithmetic Progression?

    I tried to derive a formula, and came up with this:

    . . With integers from 1 to \,n,

    . . . . \begin{array}{ccc}\text{If }n\text{ is odd, }2k+1: & k^2\text{ AP's.} \\ \\[-3mm]<br />
\text{If }n\text{ is even, }2k: & k(k-1)\text{ AP's.}<br />
\end{array}


    Hence, for n = 21,\:k = 10,\;\text{ there are }10^2 = 100 AP's.


    Therefore: . P(\text{A.P.}) \;=\;\dfrac{100}{1330} \;=\;\dfrac{10}{133}

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  10. #10
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    Quote Originally Posted by Soroban View Post
    Hello, earthboy!

    I tried to derive a formula, and came up with this:

    . . With integers from 1 to \,n,

    . . . . \begin{array}{ccc}\text{If }n\text{ is odd, }2k+1: & k^2\text{ AP's.} \\ \\[-3mm]<br />
\text{If }n\text{ is even, }2k: & k(k-1)\text{ AP's.}<br />
\end{array}

    can you give a hint on how to prove this fact???
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