Respuesta :

LammettHash
Presumably, the function is

[tex]F(x,y)e^{8x-x^2}+8y-y^2[/tex]

We have

[tex]\dfrac{\partial F}{\partial x}=(8-2x)e^{8x-x^2}[/tex]
[tex]\dfrac{\partial F}{\partial y}=8-2y[/tex]

Both partial derivatives vanish when

[tex](8-2x)e^{8x-x^2}=0\implies 8-2x=0\implies x=4[/tex]
[tex]8-2y=0\implies y=4[/tex]

so there is only one critical point [tex](4,4)[/tex]. The Hessian matrix for [tex]F(x,y)[/tex] is

[tex]\mathbf H(x,y)=\begin{bmatrix}\dfrac{\partial^2F}{\partial x^2}&\dfrac{\partial^2F}{\partial x\partial y}\\\\\dfrac{\partial^2F}{\partial y\partial x}&\dfrac{\partial^2F}{\partial y^2}\end{bmatrix}=\begin{bmatrix}e^{8x-x^2}(62-32x+4x^2)&0\\0&-2\end{bmatrix}[/tex]

At the critical point, we have

[tex]\det\mathbf H(4,4)=4e^{16}>0[/tex]
[tex]\dfrac{\partial^2F}{\partial x^2}\bigg|_{(x,y)=(4,4)}=-2e^{16}<0[/tex]

which indicates that a relative maximum occurs at [tex](4,4)[/tex], and the function takes on a maximum value of [tex]F(4,4)=16+e^{16}[/tex].

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