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Homework Help: Find the global extrema

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the global extrema of the function f(x,y)=sinxy on the closed region given by 0<x<pi and 0<y<pi. You must provide a complete analysis of the function both inside the region and on each of the boundary lines. Be sure to clearly indicate the maximum and minimum values and all places where they occur.

    2. Relevant equations
    The first and second derivates of sinxy

    3. The attempt at a solution
  2. jcsd
  3. Nov 16, 2008 #2
    This seems hard!...But I think its not!{BUT IT COULD BE HARD!}All you need to know is the second derivative test for multivariate functions!That is D=(Fxx)(Fyy)-(Fxy)^2...the D is delta!...steps in the solution would be as follows:find the first order partial derivatives of f(x,y) with respect to x [df/dx=ycosxy] and with respect to y[df/dy=xcosxy] then equate these two with zero to have solutions of extreme points (y=0,xy=pi/2 and x=0)..subsititute these values into f(x,y) to have extreme values (0,1 and 0) respectively!The SECOND DERIVATIVE test gets in handy now,to show you where there are relative minima,maxima,saddle point or indeterminate cases and to provide an equation which will give us a region in the x-y plane to define x and y values which are present in the region of extremity...take 2nd order partial deritives [Fxx=-y^2 sinxy],[Fyy=-x^2 sinxy] and mixed partial derivative [Fxy=-yx sinxy]!...I'd advise now to have a table to evaluate D <from above,beginning> with proper correspondence...Cases are if D>0,and Fxx>0 then relative minimum,D>0 and Fxx<0 then relative maximum,D<0 then saddle point and when D=0 then it is an indeterminate case!..proper evaluation will correspond extremum points (y=0,xy=pi/2 and x=0) with Delta cases ( indeterminate, [(yx)^2-(pi/2)^2] and indeterminate) respectively,leave the indeterminate cases aside and focus on the middle...(xy)^2-(pi/2)^2=D...use the case conditions..1. to have relative maximum y<0 and y>pi/2x...2. to have relative minimum then y>0 and y>pi/2x and finally to have saddle point y<pi/2x!...And if you so much care an inderterminate case at y=pi/2x..KEEP IN MIND both x and y obey the initial conditions 0<x<pi and 0<y<pi!So might as well sketch a graph to demonstrate the FEASIBLE REGION.. !...contact me through gipstud@yahoo.com for further discussions.
    Last edited: Nov 16, 2008
  4. Nov 17, 2008 #3


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    The second derivative is irrelevant here. However the most important part of any problem is that you try! ookt2c, I see no evidence that you have tried at all. What is the derivative of sin(xy)? Once you have found it what are you going to do with it?

    Perhaps the most important part of this problem is the reference to the "boundary lines". What is the boundary of the set?
  5. Nov 17, 2008 #4
    eerm..Sir I think the second derivative test is very relevant!Its a multivariate function,how else would you be able to do it,leaving aside using completing the square<which is irrelevant> or graphing of the vertical traces?...and why are you asking for just a "derivative" instead of partial derivatives..I mean this is a multivariate function! Though I agree the poster shouldve shown his approach to determine level of approaching the question!
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