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
"=(2x-4x^3-20xy^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 (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}"
"(0, 0), (0, -\\dfrac{\\sqrt{2}}{2}), (0, \\dfrac{\\sqrt{2}}{2})"
"(0, 0), ( -\\dfrac{\\sqrt{2}}{2},0), (\\dfrac{\\sqrt{2}}{2},0)"
"xy\\not=0"
This system has no solution.
Critical points:
"\\dfrac{\\partial^2 f}{\\partial x^2}=(2-12x^2-20y^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}"
"\\big( -\\dfrac{\\sqrt{2}}{2},0\\big)"
"\\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\\partial y}|_{(0,-\\sqrt{2}\/2)}=0"
"\\big( 0,0\\big)"
"\\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\\partial y}|_{(0,\\sqrt{2}\/2)}=0"
"\\big( \\dfrac{\\sqrt{2}}{2},0\\big)"
"\\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)"
Point "\\big( -\\dfrac{\\sqrt{2}}{2},0\\big)" is a saddle point.
"\\big( \\dfrac{\\sqrt{2}}{2},0\\big)"
Point "\\big( \\dfrac{\\sqrt{2}}{2},0\\big)" is a saddle point.
b)
"\\big( 0,-\\dfrac{\\sqrt{2}}{2}\\big)"
Point "\\big(0, -\\dfrac{\\sqrt{2}}{2}\\big)" is a local maximum.
"\\big( 0,\\dfrac{\\sqrt{2}}{2}\\big)"
Point "\\big(0, -\\dfrac{\\sqrt{2}}{2}\\big)" is a local maximum.
c)
"\\big( 0,0\\big)"
Point "\\big(0, 0\\big)" is a local minimum.
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