# Find the absolute minimum of f over __D__. f(x, y) = x^4 + 2y^4 ? 4yx^2 + 10 over the region D...

## Question:

Find the absolute minimum of f over D. {eq}f(x, y) = x^4 + 2y^4 ? 4yx^2 + 10 {/eq} over the region {eq}D = {(x, y) \epsilon \ R 2 \ |x ≥ -2, y ≥ 0}. {/eq} consider the points on the boundary too.

## Critical points:

If you find the critical points of a function, and we study the derivatives we can know the behavior of the function, if the function increases or decreases, the point of greatest increase is the maximum and the one of greatest decrease is the minimum.

The Hessian {eq}D(a,b) {/eq}

If {eq}D(a,b) > 0 \; \text{&} \; f_{xx} > 0 \; \Rightarrow \; \; \text{ f has a relative minimum point at (a,b, f(a,b))} \\ D(a,b) > 0 \; \text{&} \; f_{xx} < 0 \; \Rightarrow \; \; \text{ f has a relative maximum point at (a,b, f(a,b))} \\ D(a,b) < 0 \; \; \Rightarrow \; \; \text{then (a,b, f(a,b)) is a saddle point} \\ D(a,b) = 0 \Rightarrow \; \; \text{ the test is inconclusive} \\ {/eq}

We have the function

{eq}f(x, y) = x^4 + 2y^4 ? 4yx^2 + 10 \\ {/eq}

First partial derivatives:

{eq}f_x=4\,{x}^{3}-8\,xy \\ {/eq}

{eq}f_y= 8\,{y}^{3}-4\,{x}^{2} \\ {/eq}

Second partial derivatives:

{eq}f_{xx}= 12\,{x}^{2}-8\,y \\ f_{yy}= 24\,{y}^{2} \\ f_{yx}= -8\,x \\ {/eq}

Now,

{eq}f_x=0\,\, \Rightarrow \, 4\,{x}^{3}-8\,xy =0\\ f_y=0\,\, \Rightarrow \, 8\,{y}^{3}-4\,{x}^{2} =0\\ {/eq}

The possible critical points are:

{eq}(0,0) \\ (\sqrt{2}, 1) \\ {/eq}

Second derivative test for:

Point: {eq}(0,0)\\ {/eq}

{eq}f_{xx} (0,0)=0 \\ f_{yy} (0,0)= 0 \\ f_{xy} (0,0)= 0 \\ {/eq}

Hessian is given by:

{eq}\displaystyle D(a,b)= f_{xx}(a,b).f_{yy}(a,b)-[ f_{xy} (a,b) ]^{2} \\ {/eq}

So, Hessian for {eq}(0,0) {/eq} is:

{eq}\displaystyle D(0,0)= f_{xx}(0,0).f_{yy}(0,0)-[ f_{xy} (0,0) ]^{2} \\ \displaystyle D(0,0)= 0 \\ {/eq}

{eq}\begin{array} \; \; \text{(a,b)} \; & { \ f_{xx} (a,b) } & { f_{yy} (a,b) } & f_{xy} (a,b) & D(a,b) & Conclusion & (a,b, f(a,b)) \\ \hline (0,0) & 0 & 0 & 0 & 0 & the \, test \, is \, inconclusive \; & (0,0, 0) \\ (\sqrt{2}, 1) & 16 & 24 & -8\,\sqrt {2} & 256 & Relative \, minimum & (\sqrt{2}, 1, -2) \\ \end{array} \\ {/eq}

Therefore, the function has no Absolute maximum. The function has a Minimum at: {eq}(\sqrt{2}, 1, -2) \\ {/eq}