Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro Multivariate Calculus - Bruce E. Shapiro
126 LECTURE 15. CONSTRAINED OPTIMIZATION: LAGRANGE MULTIPLIERS The extremum occur at ( −1 + √ 1 , −1 + √ 1 , −1 + √ ) 2 2 2 and ( −1 − √ 1 , −1 − √ 1 , −1 − √ ) 2 2 2 The minimum is at ( f −1 + √ 1 , −1 + √ 1 , −1 + √ ) 2 2 2 ( = 2 −1 + √ 1 ) 2 ( + −1 + √ ) 2 2 2 = 2 − 4 √ 2 + 1 + 1 − 2 √ 2 + 2 = 4 − 4 √ 2 and the maximum occurs when ( f −1 − √ 1 , −1 − √ 1 , −1 − √ ) 2 2 2 ( = 2 −1 − √ 1 ) 2 ( + −1 − √ ) 2 2 2 = 2 + 4 √ 2 + 1 + 1 + 2 √ 2 + 2 = 6 + 4 √ 2 Revised December 6, 2006. Math 250, Fall 2006
Lecture 16 Double Integrals over Rectangles Recall how we defined the Riemann integral as an area in Calculus I: to find the area under a curve from a to b we partitioned the interval [a, b] into the set of numbers a = x 0 < x 1 < x 2 < · · · < x n−1 < x n = b We then chose a sequence of points, one inside each interval c i ∈ (x i−1 , x i ), x i−1 < c i < x i and approximated the area A i under the curve from x i−1 to x i with a rectangle with a base width of δ i = x i − x i−1 and height f(c i ), so that A i = δ i f(c i ) Figure 16.1: Calculation of the Riemann Sum to find the area under the curve from a = x 1 to b = x n approximates the area by a sequence of rectangles and then takes the limit as the number of rectangles becomes large. The total area under the curve from a to b was then approximated by a sequence of such rectangles, as the Riemann Sum A ≈ n∑ A i = i=1 n∑ δ i f(c i ) i=1 127
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126<br />
LECTURE 15.<br />
CONSTRAINED OPTIMIZATION: LAGRANGE<br />
MULTIPLIERS<br />
The extremum occur at<br />
(<br />
−1 + √ 1 , −1 + √ 1 , −1 + √ )<br />
2<br />
2 2<br />
and (<br />
−1 − √ 1 , −1 − √ 1 , −1 − √ )<br />
2<br />
2 2<br />
The minimum is at<br />
(<br />
f −1 + √ 1 , −1 + √ 1 , −1 + √ )<br />
2<br />
2 2<br />
(<br />
= 2 −1 + √ 1 ) 2 (<br />
+ −1 + √ ) 2<br />
2<br />
2<br />
= 2 − 4 √<br />
2<br />
+ 1 + 1 − 2 √ 2 + 2<br />
= 4 − 4 √ 2<br />
and the maximum occurs when<br />
(<br />
f −1 − √ 1 , −1 − √ 1 , −1 − √ )<br />
2<br />
2 2<br />
(<br />
= 2 −1 − √ 1 ) 2 (<br />
+ −1 − √ ) 2<br />
2<br />
2<br />
= 2 + 4 √<br />
2<br />
+ 1 + 1 + 2 √ 2 + 2<br />
= 6 + 4 √ 2 <br />
Revised December 6, 2006. Math 250, Fall 2006
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