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Unformatted text preview: , √ dC/dx = 0 when x = 3 3375 = 15, d2 C/dx2 > 0 so C is least when x = 15, y = 10. 21. Let x = length of each edge of base, y = height, k = \$/cm2 for the sides. The cost is C = (2k )(2x2 ) + (k )(4xy ) = 4k (x2 + xy ), but V = x2 y = 2000 thus y = 2000/x2 so C = 4k (x2 + 2000/x) for x > 0 dC/dx = 4k (2x − 2000/x2 ), dC/dx = 0 when √ x = 3 1000 = 10, d2 C/dx2 > 0 so C is least when x = 10, y = 20. 22. Let x and y be the dimensions shown in the ﬁgure and V the volume, then V = x2 y . The amount of material is to be 1000 ft2 , thus (area of base) + (area of sides) = 1000, x2 + 4xy = 1000, 1 1000 − x2 1000 − x2 so V = x2 = (1000x − x3 ) for y= 4x √ 4x 4 0 < x ≤ 10 10. dV 1 dV = (1000 − 3x2 ), = 0 when x = 1000/3 = 10 10/3. dx 4 dx √ 5000 10/3, 0; If x = 0, 10 10/3, 10 10 then V = 0, 3 the volume is greatest for x = 10 10/3 ft and y = 5 10/3 ft. y x x 23. Let x = height and width, y = length. The surface area is S = 2x2 + 3xy where x2 y = V , so y = V /x2 and S = 2x2 + 3V /x for x > 0; dS/dx = 4x − 3V /x2 , dS/dx = 0 when x = 3 3V /4, d2 S/dx2 > 0 so S 4 3 3V 3 3V ,y= . is minimum when x = 4 3 4 24. Let r and h be the dimensions shown in the ﬁgure, then the volume of the inscribed cylinder is V = πr2 h. But 2 h h2 r2 + = R2 thus r2 = R2 − 2 4 h h2 h3 R h 2 2 2 h=π R h− so V = π R − 4 4 dV dV 3 r = π R 2 − h2 , =0 for 0 ≤ h ≤ 2R. dh 4 dh √ √ 4π when h = 2R/ 3. If h = 0, 2R/ 3, 2R then V = 0, √ R3 , 0 so the volume is largest when 33 √ h = 2R/ 3 and r = 2/3R. Exercise Set 6.2 186 25. Let r and h be the dimensions shown in the ﬁgure, then the surface area is S = 2πrh + 2πr2 . 2 √ h But r2 + = R2 thus h = 2 R2 − r2 so h √2 S = 4πr R2 − r2 + 2πr2 for 0 ≤ r ≤ R, R h 2 4π (R2 − 2r2 ) dS dS =√ = 0 when + 4πr; dr dr R2 − r 2 r R2 − 2r2 √ = −r (i) R2 − r 2 √ R2 − 2r2 = −r R2 − r2 R4 − 4R2 r2 + 4r4 = r2 (R2 − r2 ) 5r4 − 5R2 r2 + R4 = 0 √ √ √ 5± 5 5R2 ± 25R4 − 20R4 5± 5 2 2 and using the quadratic formula r = = R,r= R, of which 10 10 10 √ √ √ 5+ 5 5+ 5 R satisﬁes (i). If r = 0, R, 0 then S = 0, (5 + 5)πR2 , 2πR2 so the surface only r = 10 10 √ √ 5+ 5 5− 5 R and, from h = 2 R2 − r2 , h = 2 R. area is greatest when r = 10 10 26. Let R and H be the radius and height of the cone, and r and h the radius and height of the cylinder (see ﬁgure), then the volume of the r H −h cylinder is V = πr 2 h. By similar triangles (see ﬁgure) = H R thus H H H h = (R − r) so V = π (R − r)r2 = π (Rr2 − r3 ) for 0 ≤ r ≤ R. R R R dV H dV H = π (2Rr − 3r2 ) = π r(2R − 3r), = 0 for 0 < r < R dr R R dr when r = 2R/3. If r = 0, 2R/3, R then V = 0, 4πR2 H/27, 0 so the 4 41 2 4πR2 H = πR H = · (volume of cone). maximum volume is 27 93 9 r H h R 27. From (13), S = 2πr2 + 2πrh. But V = πr2 h thus h = V /(πr2 ) and so S = 2πr2 + 2V /r for r > 0. dS/dr = 4πr − 2V /r2 , dS/dr = 0 if r = 3 V /(2π ). Since d2 S/dr2 = 4π + 4V /r3 > 0, the minimum surface area is achieved when r = 3 V /2π and so h = V /(πr2 ) = [V /(πr3 )]r = 2r. 28. V = πr2 h where S = 2πr2 + 2πrh so h = 29. The surface area is S = πr2 + 2πrh where V = πr2 h = 500 so h = 500/(πr2 ) and S = πr2 + 1000/r for r > 0; dS/dr = 2πr − 1000/r2 = (2πr3 − 1000)/r2 , dS/dr = 0 when r = 3 500/π , d2 S/dr2 > 0 for r > 0 so S is minimum when 500 500 500 3 500/π r = 3 500/π and h = 2 = 3 r = πr πr π (500/π ) 1 S − 2πr2 , V = (Sr − 2πr3 ) for r > 0. 2πr 2 1 d2 V dV = (S − 6πr2 ) = 0 if r = S/(6π ), = −6πr < 0 so V is maximum when dr 2 dr2 S − 2πr2 S − 2πr2 S − S/3 r = S/(6π ) and h = = r = 2r, thus the height is equal to the r= 2πr 2πr2 S/3 diameter of the base. = 3 500/π. r h 187 30. Chapter 6 The total area of material used is A = Atop + Abottom + Aside = (2r)2 + (2r)2 + 2πrh = 8r2 + 2πrh. The volume is V = πr2 h thus h = V /(πr2 ) so A = 8r2 + 2V /r for r > 0, √ dA/dr = 16r − 2V /r2 = 2(8r3 − V )/r2 , dA/dr = 0 when r = 3 V /2. This is the only critical point, √ r r π3 = = r and, for d2 A/dr2 > 0 there so the least material is used when r = 3 V /2, h V /(πr2 ) V √ πV π r 3 =. r = V /2, = h V8 8 31. Let x be the length of each side of the squares and y the height of the frame, then the volume is V = x2 y . The total length of the wire is L thus 8x +4y = L, y = (L − 8x)/4 so V = x2 (L − 8x)/4 = (Lx2 − 8x3 )/4 for 0 ≤ x ≤ L/8. dV /dx = (2Lx − 24x2 )/4, dV /dx = 0 for 0 < x < L/8 when x = L/12. If x = 0, L/12, L/8 then V = 0, L3 /1728, 0 so the volume is greatest when x = L/12 and y = L/12. 32. (a) Let x = diameter of the sphere, y = length of an edge of the cube. The combined volume is 1 (S − πx2 )1/2 V = πx3 + y 3 and the surface area is S = πx2 + 6y 2 = constant. Thus y = and 6 61/2 2 3/2 S (S − πx ) π ; for 0 ≤ x ≤ V = x3 + 6 π 63/2 √ dV π 3π π dV = x2 − 3/2 x(S − πx2 )1/2 = √ x( 6x − S − πx2 ). = 0 wh...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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