How to prove the existence of a minimum of a quadratic function of two variables? f ( x , y ) = A x 2 + 2 B x y + C y 2 + 2 D x + 2 E y + F , where A &gt; 0 and B 2 &lt; A C .Prove that a point (a,b) exists which f has a minimum.I figured out that there is no stationary point for this equation.So, Hessian Matrix seems not helpful.In my book, it says that "change quadratic part to sum of squaresbut, Can't think of any way to change it to sum of squares.Also,Why f ( a , b ) = D a + E b + F is at this minimum..?

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How to prove the existence of a minimum of a quadratic function of two variables?  f  (  x  ,  y  )  =  A      x    2    +  2  B  x  y  +  C      y    2    +  2  D  x  +  2  E  y  +  F  ,    where   A  &amp;gt;  0   and       B    2    &amp;lt;  A  C  .Prove that a point (a,b) exists which f has a minimum.I figured out that there is no stationary point for this equation.So, Hessian Matrix seems not helpful.In my book, it says that &quot;change quadratic part to sum of squaresbut, Can&#039;t think of any way to change it to sum of squares.Also,Why   f  (  a  ,  b  )  =  D  a  +  E  b  +  F is at this minimum..?

tiulowyn9v · Accepted Answer

Step 1You figured out the wrong thing, since by computing the partial derivatives we have that stationary points occur for:      (                            A                          B                                      B                          C                      )        (                            x                                      y                      )    =  −      (                            D                                      E                      )    .Step 2Since the determinant of the matrix on the LHS is positive and   A  &amp;gt;  0, there is a unique stationary point that is a global minimum for f - it is the center of the associated conic section.

Siena Erickson · Accepted Answer

Step 1Let   H  =      (                            A                          B                                      B                          C                      )    ,  g  =      (                            D                                      E                      )  , then we have   f  (  x  )  =  ⟨  x  ,  H  x  ⟩  +  2  ⟨  g  ,  x  ⟩  +  FNote that   det  H  =  A  C  −      B    2    &amp;gt;  0, so H is invertible.Show that H is positive definite (this is straightforward, but not immediate, use the fact that the eigenvalues of a symmetric real matrix are real). This shows that f is convex, hence a stationary point is a minimiser.Then                     ∂        f        (        x        )                    ∂        x              (  δ  )  =  2  (  H  x  +  g      )    T    δ. It is straightforward to find an x such that                     ∂        f        (        x        )                    ∂        x              =  0

How to prove the existence of a minimum of a quadratic function of two variables? (x, y)=Ax^2+2Bxy+Cy^2+2Dx+2Ey+F, where A>0 and B^2<AC.

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