By George W. Bluman
ISBN-10: 0387961720
ISBN-13: 9780387961729
ISBN-10: 1461211328
ISBN-13: 9781461211327
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Let x be the length dimension of the removed squares. Then the diagrams in Fig. la describe the problem. Let V( x) be the volume of the constructed box = (8 - 2x )(3 - 2x)x = 4x 3 -22x2 + 24x. D = domain of V(x): O:$; X:$; 1. ) dV dx =12x2-44x+24 = 4{x -3){3x -2); dV dx d 2V = -- = dx 2 °= x 24x - = t. ) ° if O:$; x :$; 1 = v{ X ) is concave down for O:$; x :$; 1. Hence Va) = 22°7° m3 is the maximum value of V in domain D (Fig. , the hugest such box has volume 12~ III' I. 2. Cylindrical soup cans are to be manufactured to contain a given volume V.
8b). Sa and b. __ X 47 Solved Problems By computing the slope of A(x) at some point in D, one finds that A'(x) > 0 in D. Thus the maximum value of A(x) occurs for x =10. In this case A(lO) = HI0)(90) = 1450 mzl. Again note that A(x) is a parabola; thus no calculus is necessary. (*)9. A sheet of metal is 10 ft long and 4 ft wide. It is bent lengthwise down the middle to form a V-shaped trough 10 ft long. What should be the width across the top of the trough in order that it have maximum capacity?
Remembering that h = hew), dA 1 1 dh -=-h+-wdw 2 2dw dA = 0 dw Substituting h 2 = ( ; r ~h= w2 4h ~ h2 = w 2 = ( W)2 4 2' into h2 +(;r 2( ; =4: r = 4 ~ w = ± 2/2 . Only w = 2/2 is in D. e. e. w> 2/2), then dA dw <0. • Thus the trough has maximum capacity when 1w = 2/2 ft I. Method 2 One can find an explicit formula for area A, by introducing the angle 8 (Fig. 9c). 9d and A(O) = 1wh = 4sinOcosO = 2sin20; dA dO = 4cos20; D = domain of A( 0): 0 S 0 S D: I' The aim is to maximize A( 0) over domain dA dO = 0 ~ cos20 = 0 ~ 20 = 'IT 3'lT 5'lT "2' 2' 2"" .