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Homework Help: Min value

  1. Mar 15, 2006 #1
    find the mimimum values of [tex]f(x,y)=4x^2+5y^2[/tex] on the circular disk: [tex]x^2+y^2=<1[/tex]

    i got 4 as the smallest with x=1 and y=0. This is wrong for some reason, I'm not sure why.
  2. jcsd
  3. Mar 15, 2006 #2
    1. How did you get four? what method did you use?
    2. Could you re-phase the question? From what you have said so far I assume that you have to differentiate f(x,y) and then solve simultaniously with the other equation, but I'm not too sure.
    Is that all the question says?
  4. Mar 15, 2006 #3
    To find the absolute minimum values of a continous function f on a closed bounded set D
    1) find the values of f at the critical points of f in D
    2) Find the extreme values of f on the boundary of D
    3)The smallest of the values in step 1 and 2 is the absolute minimum value.
    Do you know how to find the critical points and extreme values?
  5. Mar 16, 2006 #4

    critical points, you take the partial derivative and set that equal to zero. But for extreme values, i dont know how to get them for a boundary that is that of a circle
  6. Mar 16, 2006 #5


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    You said minimum value on the disc, not the circle. Isn't it obvious that f is never negative? Isn't it also obvious that f(0,0)= 0?

    To answer your other question: to find max or min on the boundary of a two-dimensional region, write parametric equations for the curve. For the unit circle, the simplest parametric equations are x= cos(t), y= sin(t), in which case 4x2+ 5y2= 4 cos2(t)+ 5 sin2(t)= 4+ sin2(t). However, as I said before, the minimum value is not on the boundary.
  7. Mar 16, 2006 #6
    what about a boundary of a three-dimensional region such as a sphere? I have another problem:

    Find the maximum and minimum values of f(x,y,z)=4x+4y+4z on the sphere x^2+y^2+z^2=1. How would I set up something like this?
  8. Mar 16, 2006 #7


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    You can solve that one by inspection. The function increasing fastest in the (1,1,1) direction (either by intuition or the gradient of f, which is constant), so the point on the sphere with x=y=z>0 is where f achieves its maximum.

    Alternatively, the method of Lagrange multipliers works generally.
  9. Mar 16, 2006 #8
    The minimum value of any square is 0.
  10. Mar 16, 2006 #9
    we were taught the lagrange multipliers method, but it was never shown in class how to do it with 3 variables. how would i set it up with the lagrange multipliers?
  11. Mar 16, 2006 #10


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    If you know how it's done with two variables, you can easily generalize the same procedure to 3 (or n) variables. The level surface f(x,y,z)=c has to 'touch' (be tangent to) the level surface given by the contraint g(x,y,z)=k. So [itex]\nabla f=\lambda \nabla g[/itex]. Together with [itex]g(x,y,z)=k[/itex] this gives 4 equations in 4 unknowns.
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