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15.7.Ex42-43

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

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S E C T I O N 15.7 Optimization in Several Variables (ET Section 14.7) 753 The points where g has extreme values in the interval 0 x 1 are x = 0, x = 1, and x = 36 121 . We find the 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 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 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 O A 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 36 121 , 5 11 0 . 0005. 42. Show that the rectangular box (including a top and bottom) with fixed volume V with the smallest possible surface area is a cube (Figure 21). FIGURE 21 Rectangular box with sides x , y , z . SOLUTION Step 1. Find 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 xy , hence by (1) we get S = S ( x , y ) = 2 x · V xy + y · V xy + xy = 2 V y + V x + xy = 2 V y + 2 V x + 2 xy That is, S = 2 V y + 2 V x + 2 xy Step 2. Determine the domain. The variables x and y express lengths, therefore, they must be nonnegative. Also, S is not defined if x = 0 or y = 0, therefore the domain is D = { ( x , y ) : x > 0 , y > 0 }

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