Results 1 to 7 of 7

Math Help - Could you please check my working and help me complete..thank you

  1. #1
    Member
    Joined
    Dec 2007
    Posts
    80

    Could you please check my working and help me complete..thank you

    I have a q..bin with volume pie/8m^3 made from metal, to keep costs low it should have the min possible surface area, Am^2.
    * using pie/8m^3 write an expression for height t in terms of radius r..
    Which iv done pie/8m=pier^2Xh, then , pie/8 - pier^2=h, then, pie/8pie r^2=h, then, 1/8r^2=h, is this write?
    then show this expressen as A=pier^2+pie/4r
    area of base= pier^2
    curved area side hX2pier=2pie r h
    total area pier^2+2pie r h
    pie r^2+2 pie r 1/8r^2
    and finally A=pie r ^2+ pie/4r is this right?

    then it says to find dA/r and d^2A/dr^2
    which i have done 2pier-pie/4r^2 then 2pie+pie/2r^3
    But finally i need to find the corresponding minimum values..but im not sure how too so this....
    can you tell me how please?
    Thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    "Pie?".. So, it has to do with baking pies?.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2007
    Posts
    80
    No lol the pie expression in maths , i dont know how to put it as a symbol on this
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    It's spelled 'Pi'. I had to, I'm sorry.

    I assume when you say radius, the bin is cylindrical?.

    I can tell you that the surface area is minimum when the height equals the diameter.


    We have V={\pi}r^{2}h=\frac{\pi}{8}...[1]

    The surface area, assuming the bin has a top and bottom, would be

    S=2{\pi}rh+2{\pi}r^{2}...[2]

    This is what must be minimized.

    So, we solve [1] for h andf sub into [2]:

    Solving [1] for h, we get h=\frac{1}{8r^{2}}

    That goes into the surface formula:

    2{\pi}r(\frac{1}{8r^{2}})+2{\pi}r^{2}

    Differentiate:

    S'(r)=4{\pi}r-\frac{\pi}{4r^{2}}

    Set to 0 and solve for r and we get r=\frac{2^{\frac{2}{3}}}{4}\approx{0.397}

    Plug that into h and it should be twice that.

    h=\frac{1}{8(.397)}=.794=\frac{2^{\frac{2}{3}}}{2}

    Sure enough. So, the surface area is minimum when the height equals the diameter.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2007
    Posts
    80
    The bin only has a bottom,, does this change it?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by Chez_ View Post
    The bin only has a bottom,, does this change it?
    Exterior Area becomes:
    2 \pi r h + 1 \pi r^2
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    For only a bottom: (Using Galactus's post)

    Quote Originally Posted by galactus View Post

    We have V={\pi}r^{2}h=\frac{\pi}{8}...[1]

    The surface area, assuming the bin has only a bottom:

    S=2{\pi}rh+1{\pi}r^{2}...[2]

    This is what must be minimized.

    So, we solve [1] for h and sub into [2]:

    Solving [1] for h, we get h=\frac{1}{8r^{2}}

    That goes into the surface formula:

    2{\pi}r(\frac{1}{8r^{2}})+1{\pi}r^{2}

    Differentiate:

    S'(r)=2{\pi}r - \frac{\pi}{4r^{2}}

    Set to 0 and solve for r and we get
    2 \pi r=\frac{\pi}{4 r^2}
    r^3=\frac{1}{8}
    r=\frac{1}{2}

    Plug that into h.

    h=\frac{1}{8(0.25)}=\frac{1}{2}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. check my working
    Posted in the Statistics Forum
    Replies: 1
    Last Post: July 15th 2010, 10:26 PM
  2. Can you please check my working...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 9th 2010, 03:38 PM
  3. Replies: 2
    Last Post: November 2nd 2009, 06:11 PM
  4. Please check my working out
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 8th 2009, 07:00 AM
  5. Please can someone check my working?
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 13th 2008, 05:05 AM

Search Tags


/mathhelpforum @mathhelpforum