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constrained

# constrained - CONSTRAINED EXTREMA HOMEWORK 13.9...

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CONSTRAINED EXTREMA Math21a, O. Knill HOMEWORK: 13.9: 4,12,28,24,34 CONSTRAINED EXTREMA. Given a function f ( x, y ) of two variables and a level curve g ( x, y ) = c . Find the extrema of f on the curve. You see that at places, where the gradient of f is not parallel to the gradient of g , the function f changes when we change position on the curve g = c . Therefore we must have a solution of three equations f ( x, y ) = λ g ( x, y ) , g ( x, y ) = c to the three unknowns ( x, y, λ ). (Additionally: check points with g ( x, y ) = (0 , 0)). The constant λ is called the Lagrange mul- tiplier . The equations obtained from the gradients are called Lagrange equations . EXAMPLE. To ±nd the shortest distance from the origin to the curve x 6 + 3 y 2 = 1, we extremize f ( x, y ) = x 2 + y 2 under the constraint g ( x, y ) = x 6 + 3 y 2 1 = 0. SOLUTION: f = (2 x, 2 y ) , g = (6 x 5 , 6 y ). f = λ g gives the system 2 x = λ 6 x 5 , 2 y = λ 6 y, x 6 + 3 y 2 1 = 0. We get λ = 1 / 3 , x = x 5 , so that either x = 0 or 1 or 1. From the constraint equation, we obtain y = r (1 x 6 ) / 3. So, we have the solutions (0 , ± r 1 / 3) and (1 , 0) , ( 1 , 0). To see which is the minimum, just evaluate f on each of the points. We see that (0 , ± r 1 / 3) are the minima. HIGHER DIMENSIONS. The above constrained extrema problem works also in more dimensions. For example, if f ( x, y, z ) is a function of three variables and g ( x, y, z ) = c is a surface, we solve the system of 4 equations f ( x, y, z ) = λ g ( x, y, z ) , g ( x, y, z ) = c to the 4 unknowns ( x, y, z, λ ). If we want to extremize f ( x, y, z ) under two constrants g ( x, y, z ) = c and h ( x, y, z ) = d , we have a system of 5 equations for 5 unknowns: f ( x, y, z ) = λ g ( x, y, z ) + μ h ( x, y, z ) , g ( x, y, z ) = c, h ( x, y, z ) = d EXAMPLE. Extrema of
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