Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Optimization math Problems

  1. Nov 29, 2008 #1
    1. The problem statement, all variables and given/known data

    1. A closed box of square base and volume 36 cm^3 is to be constructed and silver plated on the outside. Silver plating for the top and the base costs 40 cnets per cm^2 and silver plating for the sides costs 30 cents per cm^2. Calculat ethe cost of plating the box so that the cost is minimized.

    2. An athletics track has two "straights" of length l meters and two semi-circular ends of radius x meters. The permieter of the track is 400 m. What values of l and x produce the largest area inside the track? [see attachment #5 for diagram]

    2. Relevant equations



    3. The attempt at a solution

    1. [tex]V = l^{2}h = 36[/tex]

    [tex]SA = l^{2} + 2lh[/tex]

    [tex]h = 36/l^{2} [/tex]

    [tex]SA = l^{2} + 72l^{-1}[/tex]

    [tex]SA' = 2l - 72l^{-2} = 2l - 72/l^{2} = 0[/tex]

    [tex]2l^{3} - 72 = 0 [/tex]

    [tex]2l^{3} = 72[/tex]

    [tex]l = 3.302, w = 3.302, h = 3.302[/tex]

    [tex]C = (2lw)(.40) + 2(lh)(.30) + 2(hw)(.30) = $21.82[/tex]

    Edit: Latex makes my final cost not clear, it says $21.82.

    Answer Key: $21.60


    2. [tex]2\pix + 4x + 2l = 400[/tex]

    [tex]\pix^{2} + 2lx = A[/tex]

    [tex]2l = 400 - 2\pix - 4x[/tex]

    [tex]l = 200 - \pix - 2x[/tex]

    [tex]\pix^{2}+(400 - 2\pix - 4x)x = A[/tex]

    [tex]-\pix^{2} - 4x^{2} + 400x = A[/tex]

    [tex]A' = -2\pix - 8x + 400 = 0[/tex]

    [tex](-2\pi - 8)x = -400[/tex]

    [tex] x = 28.005, l = 56.010[/tex]

    Answer Key: x = 63.662, l = 0


    I'm not sure what I did wrong.
     

    Attached Files:

  2. jcsd
  3. Nov 29, 2008 #2
    The second one. If the Key Answer is the correct answer. Then I also I solved and achieved that answer.

    Try,

    Parameter = [tex]400 = 2 \pi r + 2l[/tex]
    Area = [tex]\pi r^{2} + 2rl[/tex]

    Note: r = x

    and work from there.

    My answer was [tex]r = \frac{200}{\pi}[/tex] and l = 0.

    I didn't have a calculator. But R seems correct.
     
    Last edited: Nov 29, 2008
  4. Nov 29, 2008 #3
    Question 1 is probably just a rounding error.
     
  5. Nov 29, 2008 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    problem 1

    I'm assuming that [itex]SA[/itex] is supposed to represent the total surface area of the box? If so, then shouldn't it be : [tex]SA = 2l^{2} + 4lh[/tex] since the box has a top and bottom and 4 sides?

    More importantly, why are you setting the derivative of the surface area equal to zero; wouldn't that minimize/maximize the surface area, not the cost?

    You should be minimizing your expression for the cost instead.

    problem 2

    why is there a [itex]4x[/itex] in your expression for the perimeter ?

    general LaTeX pointers

    to get your [itex]\pi[/itex]s to show up properly, ou need to put a space between \pi and your next letter: \pi x^2 shows up as [itex]\pi x^2[/itex] but \pix^2 shows up as [itex]\pix^2[/itex]

    to display the [itex]\$[/itex] sign use \$ instead of $
     
  6. Nov 29, 2008 #5
    Sorry. But yeah \times looks like the cross product.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook