15.7.Ex42-43

# 15.7.Ex42-43 - S E C T I O N 15.7 Optimization in Several...

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SECTION 15.7 Optimization in Several Variables (ET Section 14.7) 753 The points where g has extreme values in the interval 0 x 1a re x = 0, x = 1, and x = 36 121 .W efndthe y -coordinates oF these points From y = 1 x : x = 0 : y = 1 0 = 1 x = 1 : y = 1 1 = 0 x = 36 121 : y = 1 r 36 121 = 5 11 We conclude that the global extrema oF f on the curve y = 1 x ,0 x 1 are obtained at the points ( 0 , 1 ) , ( 1 , 0 ) , and ³ 36 121 , 5 11 ´ . Step 3. Conclusions. We examine the values oF f ( x , y ) = x 3 y 5 at the points obtained in the previous parts. The candi- dates For global extrema are f = 0, the values oF f on the segments OA and OB : f ( 0 , 1 ) = f ( 1 , 0 ) = 0 f ± 36 121 , 5 11 = ± 36 121 3 ± 5 11 5 = 0 . 0005 We conclude that the minimum value oF f ( x , y ) in the given domain is 0 and the maximum value is f ³ 36 121 , 5 11 ´ 0 . 0005. 42. Show that the rectangular box (including a top and bottom) with fxed volume V with the smallest possible surFace area is a cube (±igure 21). FIGURE 21 Rectangular box with sides x , y , z . SOLUTION Step 1. ±ind a Function to be maximized. The surFace area oF the box with sides lengths x , y , z is S = 2 ( xz + yz + xy ) (1) We express the surFace area in terms oF x and y alone using the equation V = xyz For the volume oF the box. This equation implies that z = V , hence by (1) we get S = S ( x , y ) = 2 ± x · V + y · V + = 2 ± V y + V x + = 2 V y + 2 V x + 2 That is, S = 2 V y + 2 V x + 2 Step 2. Determine the domain. The variables x and y express lengths, thereFore, they must be nonnegative. Also, S is not defned iF x = 0or y = 0, thereFore the domain is D = { ( x , y ) : x > 0 , y > 0 }

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## This note was uploaded on 02/13/2010 for the course MATH MATH 32A taught by Professor Park during the Fall '09 term at UCLA.

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15.7.Ex42-43 - S E C T I O N 15.7 Optimization in Several...

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