1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Optimization: Minimum Surface Area

  1. Jul 19, 2008 #1
    1. The problem statement, all variables and given/known data

    I need help on an optimization problem involving a hexagonal prism with no bottom or top, but the top is covered by a trihedral pyramid which has a displacement, x, such that the surface area of the object is at a minimum for a given volume. The assigned variables include:
    A= the surface area of the object
    a= the length of each side of the hexagon (a=1)
    h= the height of the prism
    x= displacement

    2. Relevant equations

    Given that A= 6(ah-1/2ax) + 3a[tex]\sqrt{3}[/tex]*[tex]\sqrt{x^2+(a^2/4)}[/tex], use calculus to find the displacement, x that yields the minimum surface area. Calculate x if a=1.


    3. The attempt at a solution

    I'm struggling trying to find a secondary equation for this optimization problem and I think I need to use an equation evolving h, but I'm not sure what.
     
  2. jcsd
  3. Jul 19, 2008 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    The volume V is to be constant, this will lead you to an equation in h and x, from which you may find an expression for h.
     
  4. Jul 20, 2008 #3
    How would I use a formula for volume since the problem works with two shapes.

    V=(3[tex]\sqrt{3}[/tex])/2 a^2*h

    h= V/ (3[tex]\sqrt{3}[/tex]/2 a^2*h

    Then, I can just plug this into the first equation, simplify, and start the calculus. Is that right?
     
  5. Jul 20, 2008 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Your TOTAL volume is the sum of the prism's volume and the pyramid's volume.

    That is the equation in h and x you are to set up!
     
  6. Jul 20, 2008 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Your total volum is therefore:
    [tex]V=\frac{3\sqrt{3}}{2}a^{2}h+\frac{1}{3}\frac{3\sqrt{3}}{2}a^{2}x\to{h}=\frac{2V}{3\sqrt{3}a^{2}}-\frac{x}{3}[/tex]
     
  7. Jul 20, 2008 #6
    So after substituting that in for h, the simplified form would be:

    A= 4V-5x[tex]\sqrt{3}[/tex]+ 9[tex]\sqrt{x^2+4}[/tex]

    [tex]\sqrt{3}[/tex]
     
  8. Jul 20, 2008 #7
    How would I take the derivative of this equation then with multiple variables?
     
  9. Jul 21, 2008 #8

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    You only have one variable, x. Remember, V is a constant!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?