91Ó°ÊÓ

The given function is$g(x,y)=\frac{1}{x^2+y^2+1}$

The function is differentiable at every point except $x^2+y^2+1=0$

Hence, the gradient of function is $∇g(x,y,z)=\frac{∂g}{∂x}\mathbf{i}+\frac{∂g}{∂y}\mathbf{j}$

$=\left(\frac{-2x}{{\left(x^2+y^2+1\right)}^2}\right)\mathbf{i}+\left(\frac{-2y}{{\left(x^2+y^2+1\right)}^2}\right)\mathbf{j}$

The gradient vanishes at critical point

$∇g(x,y)=0$

$\frac{-2x}{{\left(x^2+y^2+1\right)}^2}=0$, $\frac{-2y}{{\left(x^2+y^2+1\right)}^2}=0$

Point satisfying both equations is $(0,0)$

Hence, critical point is$(0,0)$

Second order derivative is given by

$\frac{{∂}^{2}g}{∂x^2}=-2\frac{y^2-x^2+1}{{\left(x^2+y^2+1\right)}^3},\frac{{∂}^{2}g}{∂y^2}=-2\frac{x^2-y^2+1}{{\left(x^2+y^2+1\right)}^3},\frac{{∂}^{2}g}{∂y∂x}=\frac{4xy}{{\left(x^2+y^2+1\right)}^2}$

The discriminate is given by

$H_g(x,y)=\frac{{∂}^{2}g}{∂x^2}\frac{{∂}^{2}g}{∂x^2}-{\left(\frac{{∂}^{2}g}{∂y∂x}\right)}^2$

$=\left(-2\frac{y^2-x^2+1}{{\left(x^2+y^2+1\right)}^3}\right)\left(-2\frac{x^2-y^2+1}{{\left(x^2+y^2+1\right)}^3}\right)-{\left(\frac{4xy}{{\left(x^2+y^2+1\right)}^3}\right)}^2$

$=\frac{4\left(y^2-x^2+1\right)\left(x^2-y^2+1\right)-16x^2{y}^2}{{\left(x^2+y^2+1\right)}^6}$

$=4\frac{-y^4-x^4-2x^2{y}^2+1}{{\left(x^2+y^2+1\right)}^6}$

Simplifying numerator gives $H_g(x,y)=\frac{4\left(1-x^2-y^2\right)}{{\left(x^2+y^2+1\right)}^5}$

At $\left(0,0\right)$

$H_g(0,0)=4>0,\text{and}\frac{{∂}^{2}g}{∂x^2}=-2<0,\frac{{∂}^{2}g}{∂y^2}=-2<0$

Hence local maxima is at $(0,0)$

The maximum value is$g(0,0)=1$