Answer to Question #263114 in Calculus for sai

Question #263114

Find the critical points of the function f(x, y) = (x2+5y2)e1-2x^2-2y^2 and use the second

derivative test to classify them as local maxima, local minima, or saddle point


1
Expert's answer
2021-11-09T14:51:51-0500
"\\dfrac{\\partial f}{\\partial x}=(2x-4x(x^2+5y^2))e^{1-2x^2-2y^2}"

"=(2x-4x^3-20xy^2)e^{1-2x^2-2y^2}"



"\\dfrac{\\partial f}{\\partial y}=(10y-4y(x^2+5y^2))e^{1-2x^2-2y^2}"

"=(10y-4x^2y-20y^3)e^{1-2x^2-2y^2}"

Find the critical point(s)


"\\begin{cases}\n \\dfrac{\\partial f}{\\partial x}=0 \\\\ \\\\\n \\dfrac{\\partial f}{\\partial y}=0\n\\end{cases}"

"=>\\begin{cases}\n (2x-4x(x^2+5y^2))e^{1-2x^2-2y^2}=0 \\\\ \\\\\n (10y-4y(x^2+5y^2))e^{1-2x^2-2y^2}=0\n\\end{cases}"

"=>\\begin{cases}\n 2x(1-2(x^2+5y^2))=0 \\\\ \\\\\n 2y(5-2(x^2+5y^2))=0\n\\end{cases}"


"\\begin{cases}\n x=0 \\\\ \\\\\n 5y(1-2y^2)=0\n\\end{cases}"

"(0, 0), (0, -\\dfrac{\\sqrt{2}}{2}), (0, \\dfrac{\\sqrt{2}}{2})"


"\\begin{cases}\n y=0 \\\\ \\\\\n x(1-2x^2)=0\n\\end{cases}"

"(0, 0), ( -\\dfrac{\\sqrt{2}}{2},0), (\\dfrac{\\sqrt{2}}{2},0)"



"xy\\not=0"


"\\begin{cases}\n 1-2(x^2+5y^2)=0 \\\\ \\\\\n 5-2(x^2+5y^2)=0\n\\end{cases}=>\\begin{cases}\n 2(x^2+5y^2)=1 \\\\ \\\\\n 2(x^2+5y^2)=5\n\\end{cases}"

This system has no solution.


Critical points:


"\\big( -\\dfrac{\\sqrt{2}}{2},0\\big),\\big( 0,-\\dfrac{\\sqrt{2}}{2}\\big), \\big(0, 0\\big), \\big(0,\\dfrac{\\sqrt{2}}{2}\\big),\\big(\\dfrac{\\sqrt{2}}{2},0\\big)"

"\\dfrac{\\partial^2 f}{\\partial x^2}=(2-12x^2-20y^2)e^{1-2x^2-2y^2}"




"-4x(2x-4x^3-20xy^2)e^{1-2x^2-2y^2}"


"\\dfrac{\\partial^2 f}{\\partial x\\partial y}=-40xye^{1-2x^2-2y^2}"




"-4y(2x-4x^3-20xy^2)e^{1-2x^2-2y^2}"

"\\dfrac{\\partial^2 f}{\\partial y^2}=(10-4x^2-60y^2)e^{1-2x^2-2y^2}"




"-4y(2x-4x^3-20xy^2)e^{1-2x^2-2y^2}"

"\\big( -\\dfrac{\\sqrt{2}}{2},0\\big)"


"\\dfrac{\\partial^2 f}{\\partial x^2}|_{(-\\sqrt{2}\/2,0)}=2-6-0-4+4=-4"

"\\dfrac{\\partial^2 f}{\\partial x\\partial y}|_{(-\\sqrt{2}\/2,0)}=0"

"\\dfrac{\\partial^2 f}{\\partial y^2}|_{(-\\sqrt{2}\/2,0)}=10-2=8"

"\\big( 0,-\\dfrac{\\sqrt{2}}{2}\\big)"


"\\dfrac{\\partial^2 f}{\\partial x^2}|_{(0,-\\sqrt{2}\/2)}=2-0-10-0=-8"

"\\dfrac{\\partial^2 f}{\\partial x\\partial y}|_{(0,-\\sqrt{2}\/2)}=0"


"\\dfrac{\\partial^2 f}{\\partial y^2}|_{(0,-\\sqrt{2}\/2)}=10-0-30-0=-20"

"\\big( 0,0\\big)"


"\\dfrac{\\partial^2 f}{\\partial x^2}|_{(0,0)}=2e"

"\\dfrac{\\partial^2 f}{\\partial x\\partial y}|_{(0,0)}=0"

"\\dfrac{\\partial^2 f}{\\partial y^2}|_{(0,0)}=10e"

"\\big(0,\\dfrac{\\sqrt{2}}{2}\\big)"


"\\dfrac{\\partial^2 f}{\\partial x^2}|_{(0,\\sqrt{2}\/2)}=2-0-10-0=-8"

"\\dfrac{\\partial^2 f}{\\partial x\\partial y}|_{(0,\\sqrt{2}\/2)}=0"


"\\dfrac{\\partial^2 f}{\\partial y^2}|_{(0,\\sqrt{2}\/2)}=10-0-30-0=-20"

"\\big( \\dfrac{\\sqrt{2}}{2},0\\big)"


"\\dfrac{\\partial^2 f}{\\partial x^2}|_{(\\sqrt{2}\/2,0)}=2-6-0-4+4=-4"

"\\dfrac{\\partial^2 f}{\\partial x\\partial y}|_{(\\sqrt{2}\/2,0)}=0"

"\\dfrac{\\partial^2 f}{\\partial y^2}|_{(\\sqrt{2}\/2,0)}=10-2=8"

a)

"\\big( -\\dfrac{\\sqrt{2}}{2},0\\big)"


"D=\\begin{vmatrix}\n -4 & 0 \\\\\n 0 & 8\n\\end{vmatrix}=-32<0"

Point "\\big( -\\dfrac{\\sqrt{2}}{2},0\\big)" is a saddle point.



"\\big( \\dfrac{\\sqrt{2}}{2},0\\big)"


"D=\\begin{vmatrix}\n -4 & 0 \\\\\n 0 & 8\n\\end{vmatrix}=-32<0"

Point "\\big( \\dfrac{\\sqrt{2}}{2},0\\big)" is a saddle point.


b)

"\\big( 0,-\\dfrac{\\sqrt{2}}{2}\\big)"


"D=\\begin{vmatrix}\n -8 & 0 \\\\\n 0 & -20\n\\end{vmatrix}=160>0""\\dfrac{\\partial^2 f}{\\partial x^2}|_{(0,-\\sqrt{2}\/2)}=-8<0"

Point "\\big(0, -\\dfrac{\\sqrt{2}}{2}\\big)" is a local maximum.


"\\big( 0,\\dfrac{\\sqrt{2}}{2}\\big)"


"D=\\begin{vmatrix}\n -8 & 0 \\\\\n 0 & -20\n\\end{vmatrix}=160>0""\\dfrac{\\partial^2 f}{\\partial x^2}|_{(0,\\sqrt{2}\/2)}=-8<0"

Point "\\big(0, -\\dfrac{\\sqrt{2}}{2}\\big)" is a local maximum.


c)

"\\big( 0,0\\big)"


"D=\\begin{vmatrix}\n 2e & 0 \\\\\n 0 & 10e\n\\end{vmatrix}=20e^2>0""\\dfrac{\\partial^2 f}{\\partial x^2}|_{(0,0)}=2e>0"

Point "\\big(0, 0\\big)" is a local minimum.


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