 # Find the critical points of the following multivariable function, and using the Second Derivative Test to find the maxima and minima of the equation below: a(x,y)=(8x-y)/(e^(x^2+y^2) Albarellak 2021-02-19 Answered
Find the critical points of the following multivariable function, and using the Second Derivative Test to find the maxima and minima of the equation below: $a\left(x,y\right)=\frac{8x-y}{{e}^{{x}^{2}+{y}^{2}}}$
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Differentiating the function,
$a=\frac{8x}{{e}^{{x}^{2}+{y}^{2}}-\frac{y}{{e}^{{x}^{2}+{y}^{2}}}}$
${a}_{x}=\frac{8\left[{e}^{{x}^{2}+{y}^{2}}\cdot 1-x\cdot {e}^{{x}^{2}+{y}^{2}}\cdot 2x\right]}{{\left({e}^{{x}^{2}+{y}^{2}}\right)}^{2}}$
${a}_{x}=\frac{8\left[1-2{x}^{2}\right]}{{e}^{{x}^{2}+{y}^{2}}}$
${a}_{y}=\frac{-\left[{e}^{{x}^{2}+{y}^{2}}\cdot 1-y\cdot {e}^{{x}^{2}+{y}^{2}}\cdot yx\right]}{{\left({e}^{{x}^{2}+{y}^{2}}\right)}^{2}}$
${a}_{y}=\frac{-\left[1-2{y}^{2}\right]}{{e}^{{x}^{2}+{y}^{2}}}$
For critical points, ${a}_{x}=0,{a}_{y}=0$
${a}_{x}=\frac{8\left[1-2{x}^{2}\right]}{{\left({e}^{{x}^{2}+{y}^{2}}\right)}^{2}}=0$
$1-2{x}^{2}=0$
$2{x}^{2}=1$
$x=±\frac{1}{\sqrt{2}}$
${a}_{y}=\frac{-\left[1-2{y}^{2}\right]}{{\left({e}^{{x}^{2}+{y}^{2}}\right)}^{2}}=0$
$1-2\frac{y}{62}=0$
$2{y}^{2}=1$
$y=±\frac{1}{\sqrt{2}}$
The critical points are $\left(±\frac{1}{\sqrt{2}},±\frac{1}{\sqrt{2}}\right)$
${a}_{x}=\frac{8\left(1-2{x}^{2}\right)}{{e}^{{x}^{2}+{y}^{2}}}$
${a}_{×}=\frac{8\left(-6x+4{x}^{3}\right)}{{e}^{{x}^{2}+{y}^{2}}}$
${a}_{y}=\frac{-\left(1-2{y}^{2}\right)}{{e}^{{x}^{2}+{y}^{}}}$