# very easy 3 probs ...

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• Dec 22nd 2010, 09:14 PM
earthboy
very easy 3 probs ...
as i am new to calculus & probability,pls show how to do these problems(may be very trivial(Worried))

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 $\displaystyle y= x^{2}$and$\displaystyle y=x+2$

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
• Dec 22nd 2010, 09:17 PM
dwsmith
2
$\displaystyle \displaystyle \int_{intersection}^{intersection}[f(x)-g(x)]dx$
• Dec 22nd 2010, 09:20 PM
dwsmith
Quote:

Originally Posted by earthboy

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?
• Dec 22nd 2010, 11:03 PM
BAdhi
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 $\displaystyle \frac{dV}{dh}=0$

take $\displaystyle \frac{d^2V}{dh^2}$. If for each calculated value for h, you'll get a minus for $\displaystyle \frac{d^2V}{dh^2}$ it means V is maximum for that value for that h
• Dec 22nd 2010, 11:55 PM
earthboy
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
• Dec 22nd 2010, 11:58 PM
CaptainBlack
Quote:

Originally Posted by dwsmith
The tickets have numbers so why are we looking for letters?

A.P. <=> Arithmetic Progression

CB
• Dec 23rd 2010, 12:22 AM
BAdhi
substitute $\displaystyle g(x)=0$,$\displaystyle f(x)=x^2$ tha'll give you the area under the curve
• Dec 23rd 2010, 01:42 AM
HallsofIvy
Quote:

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

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 $\pi r^2(h(1- r/R))= \pi h (r^2- r^3/R)$. That will have maximum volume when its derivative is 0- $\pi r^2h (2r- 3r^2/R)= 0$. $r(2- 3r/)= 0$ 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.
• Dec 23rd 2010, 04:21 AM
Soroban
Hello, earthboy!

Problem 3 is intriguing.

Quote:

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: .$\displaystyle \displaystyle _{21}C_3 \:=\:{21\choose3} \:=\:\frac{21!}{3!\,18!} \:=\:1330\text{ possible outcomes.}$

How many of them are in Arithmetic Progression?

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

. . With integers from 1 to $\displaystyle \,n$,

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

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

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

• Dec 23rd 2010, 04:40 AM
earthboy
Quote:

Originally Posted by Soroban
Hello, earthboy!

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

. . With integers from 1 to $\displaystyle \,n$,

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

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