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Homework Help: Maximum area ratios

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data
    A Norman window has the shape of a rectangle with a semicircular top. Assume that the semicircular portion of the peripmeter is three times as costly to build per metre as the straight edges. For a given area, what ratio of heigh to radius would minimize the cost.

    2. Relevant equations

    A diagram. r is radius, h is height.

    3. The attempt at a solution

    [tex]P = (\pi r)+ 2r+ h + h[/tex]

    I'm not sure if this is right. It is asking for area so I should use [tex] S.A = (\pi r^2)/2 + 2rh[/tex]?

    I'm just having trouble setting up the equation.
    Last edited: Feb 6, 2010
  2. jcsd
  3. Feb 6, 2010 #2


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    Hi fghtffyrdmns! :smile:

    (have a pi: π and try using the X2 tag just above the Reply box :wink:)
    πr2/2 + 2rh is correct for the area.

    But when they say "as costly to build per metre", I think they mean you have to take the perimeter of the semi-circle. :wink:
  4. Feb 6, 2010 #3
    Would the perimieter be just [tex]\pi r[/tex]?
  5. Feb 6, 2010 #4


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    (what happened to that π i gave you? :confused:)

    Yup! :smile:
  6. Feb 6, 2010 #5
    So I would use[tex] P = (\pi r)+ 2r+ 2h[/tex] to create the cost function of [tex] C(r) = 3(\pi r)+ 2r+ 2h[/tex]?
  7. Feb 6, 2010 #6
    Moderator's note: giving complete solutions to homework problems is against our forum guidelines.

    (Solution deleted.)
    Last edited by a moderator: Feb 6, 2010
  8. Feb 6, 2010 #7
    Your answer is correct: according to the text book. I am confused though for where the c comes from.
  9. Feb 6, 2010 #8


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    The problem does not specify the cost per unit length of the window, so a truly general expression for the cost would include this (unknown) parameter. kyp4 has denoted this as c.

    As it turns out, the solution to this problem does not depend on the value of c. So setting up the problem as if c is 1 dollar per meter (as you did) works fine.
  10. Feb 6, 2010 #9

    so I have [tex] P = (\pi r)+ 2r+ 2h [/tex] and [tex] S.A = (\pi r^2)/2 + 2rh [/tex].

    From this, I can solve for h in the area function - as it asks in terms of area.

    I get [tex] h =\frac{A}{2r} - \frac{\pi r}{4}[/tex]

    Knowing that the perimeter of the semi circle is 3x as costly, you get [tex]C(r) = 3(\pi r)+ 2r+ 2h[/tex] Substituting h into the cost function, you get [tex] C(r) = \frac{5\pi r}{2} + 2r + \frac{A}{r}[/tex]

    Now, do I determine the derivative and solve for r? After determining r, I substitute into the formula for h which then I can solve h/r?
  11. Feb 6, 2010 #10
    I apologize for revealing the complete solution. I am new to the forums and did not realize this was against forum policy. Makes sense though.

    While setting [tex]c=1[/tex] works out, it is more proper to consider it as some arbitrary variable [tex]c[/tex] and it will disappear if it turns out not to be relevant as in this case. This will guard you against the case in which [tex]c[/tex] turns out to be crucial to solving the problem, which will become evident if you end up with the variable in your resulting equations.

    That being said, the procedure you described at the end of you last post fghtffyrdmns, should lead you to the same result as I got (whose correctness I am still not certain :redface:).
  12. Feb 6, 2010 #11
    [tex] C(r) = \frac{5\pi r}{2} + 2r + \frac{A}{r} [/tex]

    [tex] C'(r) = \frac{5\pi}{2} + 2 - \frac{A}{r^2}[/tex]

    I got [tex]r = \sqrt{\frac{A}{\frac{5\pi}{2} + 2}}[/tex]

    Should I simply this or can I sub it into the equation?

    [tex] h =\frac{A}{2}{\sqrt{\frac{\frac{5\pi}{2} + 2}{A}} - \frac{\pi}{4}\sqrt{\frac{A}{\frac{5\pi}{2} + 2}[/tex]

    Now, I do not know how to solve the h/r :/. The book comes out with [tex]{\pi}+1[/tex].

    Kyp, you are correct about not putting the c=1 constant in this - however, I do not know how you did it.
    Last edited: Feb 6, 2010
  13. Feb 6, 2010 #12
    Regarding the arbitrary variable [tex]c[/tex], it appears in [tex]C(r)[/tex] as a coefficient of all the terms, that is

    C(r) = \frac{5\pi cr}{2} + 2cr + \frac{Ac}{r}

    and is canceled out when solving for [tex]r[/tex] after setting the derivative to zero, arriving at the same result you got. Try it. You've got the correct results for [tex]r[/tex] and [tex]h[/tex] and have simplified them as far as they can be simplified. Simply substitute the expressions you got above into [tex]h/r[/tex] and you'll find that the square roots cancel and combine to get you the correct result.
  14. Feb 6, 2010 #13
    See, this is where I am having a problem.

    I do not know how to solve h/r. Do I have to rationalize? Or did I make a mistake with my h value so the roots do not cancel?
  15. Feb 6, 2010 #14
    You simply substitute:

    \frac{h}{r} = \frac{1}{\sqrt{\frac{A}{\frac{5\pi}{2} + 2}}}\left[\frac{A}{2}{\sqrt{\frac{\frac{5\pi}{2} + 2}{A}} - \frac{\pi}{4}\sqrt{\frac{A}{\frac{5\pi}{2} + 2}\right] = \frac{A}{2}\frac{\left(\frac{5\pi}{2}+2\right)}{A}-\frac{\pi}{4} = \pi + 1
  16. Feb 6, 2010 #15
    See, how come the square roots disappear on both of them? The algebra is confusing me.

    Edit: I figured it out how you get the [tex]\pi + 1[/tex] after you do the first part. It's the radicals that are confusing me, however.
    Last edited: Feb 7, 2010
  17. Feb 7, 2010 #16
    Yes the radicals can make it confusing. Is it clear that

    \frac{\sqrt{\frac{\frac{5\pi}{2} + 2}{A}}}{\sqrt{\frac{A}{\frac{5\pi}{2} + 2}}} = \sqrt{\frac{\frac{5\pi}{2} + 2}{A}}\sqrt{\frac{\frac{5\pi}{2} + 2}{A}} = \frac{\frac{5\pi}{2} + 2}{A}


    \frac{\sqrt{\frac{A}{\frac{5\pi}{2} + 2}}}{\sqrt{\frac{A}{\frac{5\pi}{2} + 2}}} = 1

    ? If so, the rest should follow.
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