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Need Help on max&min problems

  1. Nov 4, 2004 #1
    Hi Everybody

    I have two max and min word questions,I didnt understand well how to solve this type question,thats why, I would like you to help me and show me how to solve these questions.I know some begining steps and know all solution way.But the problem is I dont know how to use or do those steps.Could you show me but not only show also teach me with explaining the steps sothat I can see what you did.I will have a small test tomorrow, and there will be afew min max questions.I want a learn how to those and after you show me, I will work on similar questions like them.
    I will wait your responde and helps...

    Here are two questions.

    1)My first question is:
    A silo has a hemisphrerical roof,is cylindrical sided and has a circular floor, all made of steel. find the dimensions of a silo with a volume of 755 cubic meters that uses the least amount of steel to build. justify your answer with calculus. so it is asking minimize surface area.

    note:there is a picture of a silo but there isnt anything on the picture except there is r on the top and on the bottom.also there is a h left side of silo.

    My approach is=
    So the two parameters are the height, h, of the silo, and
    the radius, r, of the silo. The volume will be the volume
    of the cylinder plus the volume of the hemisphere:
    V = (2/3)pi r^3 + pi r^2 h
    So V is given in the problem, allowing me to solve
    for h in terms of r.

    The total area is the sum of the area of the hemisphere and
    the area of the cylindrical part:

    A = 2pi r^2 + 2pi rh
    Substitute for h in the area equation, then take the
    derivative to find dA/dr. Set it to zero and solve for r.
    Then back-substitute to find h.

    2)Second question is:
    A rectangular garden laid out along your neighbor's lot line contains 432 square meters area.it is to be fenced on all sides. If the neighbor pays for half the cost of the shared fence and fencing costs $35 per meter. what should the dimensions of the garden be so that YOUR COST is a minimum? justify your answer with calculus. so it is asking minimize cost of fence ,
    not minimum perimeter.

    note:there is a rectangular picture. top side of rectangular is your fence,
    left side is shared fence, right site is your fence and there is y and at the bottom your fence and there is x.

    My approach is=
    The parameters are length, L, and width, W. Let W be
    along the shared lot-line.

    A = LW

    Where area, A, is given in the problem. From this I think I can find L in terms of W, or vice versa but I dont know how to do it right:)

    C = 35 (2L + 1.5W)

    Thank you very much for your helps.
    Last edited: Nov 5, 2004
  2. jcsd
  3. Nov 4, 2004 #2


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    On a differentiable function on a conitinuous interval, maxima and minima can occur at:
    1. Locations where the derivative is zero
    2. The endpoints of the interval

    You could just find all of the locations where a maximum or minimum could occur and check them.
  4. Nov 4, 2004 #3
    I have just learned these type questions today thats why I really dont know how to do.
    I would be so happy if you showed me the steps and put some little notes what you did there. :frown:

    Plz anybody? :frown:
    Last edited: Nov 5, 2004
  5. Nov 5, 2004 #4


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    Reading what you've done above, it seems to me you've understood the meaning of that calculus. I have not found any reason for your lack of trust.

    Anyway, you can try with this problem. It's a typical engineering problem where one have to minimize the surface of a cylindrical reservoir covered by two solid surfaces (rounded) at the top and bottom. The reservoir has a height h and a radius R. Given some volume V, try to find R(h) such that the surface exposed to an external environment is minimum.
  6. Nov 5, 2004 #5


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    Okay except that you haven't included the area of the floor. You need and additional pi r^2

    Yes, and since you were told that A= LW= 432, you know that L= 432/W and want to minimize 35(864/W+ 1.5W) or, conversely, that W= 432/L and want to minimize 35(2L+ 648/L).

    It might help to write them as 35(864W-1+ 1.5W) or
    35(2L+ 648L-1).
  7. Nov 5, 2004 #6
    I did the first one , could you guys check?

    Acube(r, h) = 2∏rh+2∏r2
    = ∏r(2h+r)

    V = ∏r2h = 755 ---> h = 755/∏r2

    Acube(r) = ∏r[(1510/∏r2)+r]

    A’(r) = [(2∏r3-1510)/(r2)]

    A’(r) = [(2∏r3-1510)/(r2)] = 0

    r3 = 755/∏ ----> r = 6,2 m

    h = 755/∏r2
    h = 6,25 m
  8. Nov 5, 2004 #7
    This is the second one, Could you check this too?

    xy = 432
    y = 432/x
    Cost = 35(2x+1.5y)

    C = 35(2x+(1.5*432/x))
    C' = 70-(22680/x^2) = 0

    x = 18

    432/18 = 24

    C= 35 (2x+1.5y)
    = 35 (2.18+1.5*24)
    = 2520$
  9. Nov 5, 2004 #8


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    V = (2/3)pi r^3 + pi r^2 h
    So, solving for h:
    [tex]h=\frac{V-(2/3)\pi r^3}{\pi r^2}[/tex]

    plug this in the area (without the floor, do the correct one with the floor area yourself):
    [tex]A = 2\pi r^2 + 2\pi rh=2\pi r^2 + 2\pi r\frac{V-(2/3)\pi r^3}{\pi r^2}[/tex]
    rewrite it with a bit of algebra:
    [tex]A=\frac{2}{3}\pi r^2+\frac{2V}{r}[/tex]
    taking the derivative and setting it equal to zero:
    [tex]A'=\frac{\frac{4}{3} \pi r^3-2V}{r^2}=0[/tex]
    gives [itex]r=(\frac{6V}{4\pi})^{1/3}[/itex]
    Notice that A'<0 if r is smaller than that critical value and A'(r)>0 if r is greater than that critical value. So you have an absolute minimum.
    Since V=755 m^3, r=7.12 m.
    then plug r back into:
    [tex]h=\frac{V-(2/3)\pi r^3}{\pi r^2}=[/tex]
    to find h.
    Just like you said in your first post.

    (Surprise: h=0.)

    The fence problem is correct as far as I can tell.
    For completeness though, you should check and add that x=18 really gives a minimum C, not merely a point where the derivative is zero.
    Last edited: Nov 5, 2004
  10. Nov 5, 2004 #9
    Thank you very much galileo ,you are a great person.
    Last edited: Nov 5, 2004
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