A simple question about extreme values

Replies

• 

  Ramani Aswath

  The function can be rewritten as x^4 + y^4 ≤ 2(x - y)^2
  and simplified to: x^4 + y^4 ≤ 2(x^2 +y^2)-4xy
  Even powers of x and y are always positive. The fourth power rapidly increases with increasing values of x and y above 1. For the inequality to hold it is necessary that x and y cannot both be +ve or -ve.
  Inspection shows that the limits for the inequality is x = a value -2 to +2, simultaneously y = a value +2 to -2.
• 

  Aruwin

  bioramani

  The function can be rewritten as x^4 + y^4 ≤ 2(x - y)^2
  and simplified to: x^4 + y^4 ≤ 2(x^2 +y^2)-4xy
  Even powers of x and y are always positive. The fourth power rapidly increases with increasing values of x and y above 1. For the inequality to hold it is necessary that x and y cannot both be +ve or -ve.
  Inspection shows that the limits for the inequality is x = a value -2 to +2, simultaneously y = a value +2 to -2.

  OK,another question related to finding the extrema.
  Given function f(x,y) = x^4 + y^4 - 2(x - y)^2,
  where f(x,y)≤ 0.
  For all values of (x,y) near (0,0),
  f(x,y)≥f(0,0) → At (0,0) f(x,y) has a local minimum
  f(x,y)≤f(0,0) → At (0,0) f(x,y) has a local maximum
  OK, so how do I know if f(x,y) is greater or lesser than f(0,0)??
  
  I am going to apply what I think and please correct me if I am wrong.
  
  The function at O is -2(x-y)^2　after expanding the function using taylor expansion.
  Substituting values of x and y with 0,
  so f(0,0) = 0 and
  substituting values of x and y with 1 and 2,
  so f(1,2) = -2 and
  thus, f(1,2) is lower than f(0,0). Generalizing it, we have f(x,y) lower than or equals to(when we substitute with 0) f(0,0) and so, f(x,y) has a local maximum value of 0 at (0,0).