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Unformatted text preview: Maxima/minima with constraints • Very often we want to find maxima/minima but subject to some constraint • Example: A wire is bent to a shape y = 1 x 2 . If a string is stretched from the origin to the wire, at what point along the wire is the length of the string minimized? We want to minimize d 2 = x 2 + y 2 . We can eliminate the y 2 using y 2 = (1 x 2 ) 2 , so we minimize the function f ( x ) = x 2 + (1 x 2 ) 2 = x 4 x 2 + 1 • We find df dx = 4 x 3 2 x . The minimum occurs at 2 x 2 1 = 0 so x = ± p 1 / 2. • We also find a local maxima at x = 0. Check with second derivatives! Patrick K. Schelling Introduction to Theoretical Methods Another approach... • Let’s do the same problem, but starting from f ( x , y ) = x 2 + y 2 , and then the differential df = 2 xdx + 2 ydy • Or we can write as, df dx = 2 x + 2 y dy dx • Then we can obtain dy dx from the equation of constraint y = 1 x 2 dy dx = 2 x Patrick K. Schelling Introduction to Theoretical Methods Another approach continued... • Then we substitute into df dx = 0, df dx = 2 x 4 xy = 0 • We can also solve df = 0 since dx is arbitrary, so 2 x 4 xy = 0 • We still get 2 x 4 x (1 x 2 ) = 0 so that x = ± q 1 2 or x = 0 as before Patrick K. Schelling Introduction to Theoretical Methods Method of Lagrange multipliers • The approaches above work, either by substituting and eliminating a variable, or by finding dy dx • However, these approaches can often lead to inconvenient algrebra • We note that we can write the constraint φ ( x , y ) = constant (sometimes we write it so that φ ( x , y ) = 0) • Then we have, for minimization of f ( x , y ) with constraint φ (...
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 Spring '03
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 Optimization, Elementary algebra, Partial differential equation, Laplace operator, Patrick K. Schelling

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