The solution of v x 0 in the domain of v is x 10

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The solution of V ( x ) = 0 in the domain of V is x = 10. Because V (10) = 1000 > V (1) = 149 . 5 > V ( 10 3 ) = 0, this shows that x = 10 maximizes V and that the maximum value of V is 1000 cm 3 . 3.6.18: Let x denote the radius of the cylinder and y its height. Then its total surface area is πx 2 +2 πxy = 300 π , so x 2 + 2 xy = 300. We are to maximize its volume V = πx 2 y . Because y = 300 x 2 2 x , it follows that V = V ( x ) = π 2 (300 x x 3 ) , 0 x 10 3 . It is then easy to show that x = 10 maximizes V ( x ), that y = x = 10 as well, and thus that the maximum possible volume of the can is 1000 π in. 3 3.6.19: Let x be the length of the edge of each of the twelve small squares. Then each of the three cross- shaped pieces will form boxes with base length 1 2 x and height x , so each of the three will have volume x (1 2 x ) 2 . Both of the two cubical boxes will have edge x and thus volume x 3 . So the total volume of all five boxes will be V ( x ) = 3 x (1 2 x ) 2 + 2 x 3 = 14 x 3 12 x 2 + 3 x, 0 x 1 2 . Now V ( x ) = 42 x 2 24 x + 3; V ( x ) = 0 when 14 x 2 8 x 1 = 0. The quadratic formula gives the two solutions x = 1 14 ( 4 ± 2 ) . These are approximately 0 . 3867 and 0 . 1847, and both lie in the domain of V . Finally, V (0) = 0, V (0 . 1847) 0 . 2329, V (0 . 3867) 0 . 1752, and V (0 . 5) = 0 . 25. Therefore, to maximize V , one must cut each of the three large squares into four smaller squares of side length 1 2 each and form the resulting twelve squares into two cubes. At maximum volume there will be only two boxes, not five. 3.6.20: Let x be the length of each edge of the square base of the box and let h denote its height. Then its volume is V = x 2 h . The total cost of the box is \$144, hence 4 xh + x 2 + 2 x 2 = 144 and thus h = 144 3 x 2 4 x . Therefore V = V ( x ) = x 4 ( 144 3 x 2 ) = 36 x 3 4 x 3 . 172
The natural domain of V is the open interval ( 0 , 4 3 ) , but we may adjoin the endpoints as usual to obtain a closed interval. Also V ( x ) = 36 9 4 x 2 , so V ( x ) always exists and is zero only at x = 4 (reject the other root x = 4). Finally, V ( x ) = 0 at the endpoints of its domain, so V (4) = 96 (ft 3 ) is the maximum volume of such a box. The dimensions of the largest box are 4 ft square on the base by 6 ft high. 3.6.21: Let x denote the edge length of one square and y that of the other. Then 4 x +4 y = 80, so y = 20 x . The total area of the two squares is A = x 2 + y 2 , so A = A ( x ) = x 2 + (20 x ) 2 = 2 x 2 40 x + 400 , with domain (0 , 20); adjoin the endpoints as usual. Then A ( x ) = 4 x 40, which always exists and which vanishes when x = 10. Now A (0) = 400 = A (20), whereas A (10) = 200. So to minimize the total area of the two squares, make two equal squares. To maximize it, make only one square. 3.6.22: Let r be the radius of the circle and x the edge of the square. We are to maximize total area A = πr 2 + x 2 given the side condition 2 πr + 4 x = 100. From the last equation we infer that x = 100 2 πr 4 = 50 πr 2 . So A = A ( r ) = πr 2 + 1 4 (50 πr ) 2 = π + 1 4 π 2 r 2 25 πr + 625 for 0 r 50 (because x 0). Now A ( r ) = 2 π + 1 4 π 2 r 25 π ; A ( r ) = 0 when r = 25 2 + π 2 = 50 π + 4 ; that is, when r 7. Finally, A (0) = 625 , A 50 π 795 . 77 and A 50 π + 4 350 . 06 .