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!

Minimum value

  1. Jul 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A rectangular building is being designed to minimize heat loss. The east and west walls lose heat at a rate of 10 units/m^2 per day, the north and south walls at a rate of 8 units/m^2 per day, the floor at a rate of 1 unit/m^2 per day, and the roof at a rate of 5 units/m^2 per day. Each wall must be at least 30 m long, the height must be at least 4 m, and the volume must be exactly 4000 m^3.

    Find the dimensions that minimize heat loss. (check the critical pts and boundary pts)

    3. The attempt at a solution
    Since Length x Width=120, the height must be 33 1/3 m.

    I'm having trouble starting with the equation for heat loss. I so far I have 2(10)x + 2(8)y + (5+1)z.

    I'm not sure about the z part. The floor loses by a factor of 1, and the roof by a factor of 5. Since they both must have equal area, I figured they must lose at the same rate--that's why I added the 2 numbers together. Can anyone explain it better?
  2. jcsd
  3. Jul 8, 2008 #2


    User Avatar
    Homework Helper

    Well for some reason you're not using the notation x,y,z to represent the dimensions directly, instead in your formulation they represent surface areas of the walls. Why not just use them directly?

    And yes it's ok to group the floor and roof areas together since their areas are the same. Just find the critical pts and boundary values as the question said. Alternatively you could use the method of Lagrange multipliers if you've learned it already.
  4. Jul 9, 2008 #3
    So are you saying instead of just 20x I should have 20x^2?

    The way my equation is now, I just get real numbers for my partial derivatives, so I can't even find any critical pts or plug in the boundary points...
  5. Jul 9, 2008 #4


    User Avatar
    Homework Helper

    No, 20x^2 is only if the walls are square, which they do not have to be. Let the height be z, the length be y and the width be x. Volume = xyz = 4000.

    To set up the total heat loss equation you have to decide which walls correspond to the compass bearings east, west, north, south and their respective surface areas. It's arbitrary but you must be consistent in this choice throughout the question.

    And as for your earlier attempt, where does it say that length x width = 120? Remember that you are given the minimum dimensions but not the maximum ones.
  6. Jul 10, 2008 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    What tells you that length times width= 120?

    I have no idea if this is correct because you haven't said what x, y, and z are! Are they areas? If you find x, y, and z, how will that help you answer the question?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Minimum value