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Use Lagrange multipliers to find any extrema of the function subject to the constraint x^2 + y^2 ? 1. f(x, y) = e^{-xy/4} minimum f( ) = (smaller x-value) minimum f( ) = (larger x-value) maximum f( ) = (smaller x-value) maximum f( ) = (larger x-value)

          Use Lagrange multipliers to find any extrema of the function subject to the
constraint x^2 + y^2 ? 1.
f(x, y) = e^{-xy/4}
minimum f( ) = (smaller x-value)
minimum f( ) = (larger x-value)
maximum f( ) = (smaller x-value)
maximum f( ) = (larger x-value)
        
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Use Lagrange multipliers to find any extrema of the function subject to the
constraint x^2 + y^2 ? 1.
f(x, y) = e^-xy/4
minimum f( ) = (smaller x-value)
minimum f( ) = (larger x-value)
maximum f( ) = (smaller x-value)
maximum f( ) = (larger x-value)

Added by Caleb B.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Use Lagrange multipliers to find any extrema of the function subject to the constraint x^2 + y^2 ≤ 1. f(x, y) = e^(-xy/4) minimum f( ) = (smaller x-value) minimum f( ) = (larger x-value) maximum f( ) = (smaller x-value) maximum f( ) = (larger x-value)
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Transcript

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00:01 We have the function xy which is equal to e to the power minus xy over 4 and the constraints as gxy is equal to x square plus y square minus 1 less than and equal to 0.
00:12 Now using the lagrange's multipliers, we let f of x minus lambda g of x to be equal to 0 and f of y minus lambda gy is equal to 0 for x and y minus lambda gy is equal to 0.
00:33 So we get the value of lambda is equal to negative y over 8x, e to the bar minus xy over 4, which will be further deduced to in terms of x, x over 8y, e to the power minus xy over 4.
00:57 Now if we put y square in the constraint, so x equal to 1 over root 2, which further implies that y is equal to plus minus of 1 over under the root of 2...
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