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Unformatted text preview: en x = 0, or when dx 2 dx 6 26 √ √ S S S S ,x= . If x = 0, , , then 6x = S − πx2 , 6x2 = S − πx2 , x2 = 6+π 6+π 6+π π S 3/2 S 3/2 S 3/2 S S , √ so that V is smallest when x = , and hence when y = , V = 3/2 , √ 6+π 6+π 6 6 6+π 6 π thus x = y . (b) From part (a), the sum of the volumes is greatest when there is no cube. 33. 34. Let h and r be the dimensions shown in the ﬁgure, then the volume 1 is V = πr2 h. But r2 + h2 = L2 thus r2 = L2 − h2 so 3 1 1 V = π (L2 − h2 )h = π (L2 h − h3 ) for 0 ≤ h ≤ L. 3 3 √ √ dV 1 dV 2 2 = π (L − 3h ). = 0 when h = L/ 3. If h = 0, L/ 3, 0 dh 3 dh 2π 3 then V = 0, √ L , 0 so the volume is as large as possible when 93 √ h = L/ 3 and r = 2/3L. Let r and h be the radius and height of the cone (see ﬁgure). The slant height of any such cone will be R, the radius of the circular sheet. Refer to the solution of Exercise 33 to ﬁnd that the largest 2π volume is √ R3 . 93 h L r h R r Exercise Set 6.2 35. 36. 37. 38. 39. 188 √ The area of the paper is A = πrL = πr r2 + h2 , but 1 V = πr2 h = 10 thus h = 30/(πr2 ) so A = πr r2 + 900/(π 2 r4 ). 3 To simplify the computations let S = A2 , 900 900 S = π 2 r2 r2 + 2 4 = π 2 r4 + 2 for r > 0, πr r 1800 4(π 2 r6 − 450) dS 23 = 4π r − 3 = , dS/dr = 0 when dr r r3 r = 6 450/π 2 , d2 S/dr2 > 0, so S and hence A is least when 30 3 2 π /450. r = 6 450/π 2 , h = π 1 hb. By similar triangles (see ﬁgure) 2 R b/2 2Rh Rh2 =√ ,b= √ so A = √ for h > 2R, h h2 − 2Rh h2 − 2Rh h2 − 2Rh 2 Rh (h − 3R) dA dA =2 = 0 for h > 2R when h = 3R, by the ﬁrst , dh (h − 2Rh)3/2 dh derivative test A is minimum when h = 3R. If h = 3R then √ b = 2 3R (the triangle is equilateral). r The area of the triangle is A = 1 The volume of the cone is V = πr2 h. By similar triangles (see 3 R r Rh ﬁgure) = √ ,r=√ so 2 − 2Rh 2 − 2Rh h h h 1 h3 h2 1 = πR2 for h > 2R, V = πR2 2 3 h − 2Rh 3 h − 2R 1 dV h(h − 4R) dV = πR2 = 0 for h > 2R when h = 4R, by the , dh 3 (h − 2R)2 dh ﬁrst √ derivative test V is minimum when h = 4R. If h = 4R then r = 2R. h−R dA/dθ = 0 for 0 < θ < π/2 when cos θ = 1/2, θ = π/3. If √ θ = 0, π/3, π/2 then A = 0, 75 3/4, 25 so the cross-sectional area is greatest when θ = π/3. h2 − 2Rh h R R b /2 b h−R h2 − 2Rh h R R r The area is (see ﬁgure) 1 A = (2 sin θ)(4 + 4 cos θ) 2 = 4(sin θ + sin θ cos θ) for 0 ≤ θ ≤ π/2; dA/dθ = 4(cos θ − sin2 θ + cos2 θ) = 4(cos θ − [1 − cos2 θ] + cos2 θ) = 4(2 cos2 θ + cos θ − 1) = 4(2 cos θ − 1)(cos θ + 1) dA/dθ = 0 when θ = π/3 for 0 < θ < π/2. If θ = 0, π/3, π/2 then √ √ A = 0, 3 3, 4 so the maximum area is 3 3. Let b and h be the dimensions shown in the ﬁgure, then the 1 cross-sectional area is A = h(5 + b). But h = 5 sin θ and 2 5 b = 5 + 2(5 cos θ) = 5 + 10 cos θ so A = sin θ(10 + 10 cos θ) 2 = 25 sin θ(1 + cos θ) for 0 ≤ θ ≤ π/2. dA/dθ = −25 sin2 θ + 25 cos θ(1 + cos θ) = 25(− sin2 θ + cos θ + cos2 θ) = 25(−1 + cos2 θ + cos θ + cos2 θ) = 25(2 cos2 θ + cos θ − 1) = 25(2 cos θ − 1)(cos θ + 1). L h 4 cos θ 2 sin θ 2 θ 4 5 cos θ 5 2 cos θ b h = 5 sin θ θ 5 5 189 Chapter 6 , k the constant of proportionality. If h is the height of the lamp above the table then √ r2 − 2h2 h h dI dI cos φ = h/ and = h2 + r2 so I = k 3 = k 2 =k 2 =0 for h > 0, , 2 )3/2 2 )5/2 dh dh (h + r (h + r √ √ when h = r/ 2, by the ﬁrst derivative test I is maximum when h = r/ 2. 41. Let L, L1 , and L2 be as shown in the ﬁgure, then L = L1 + L2 = 8 csc θ + sec θ, dL = −8 csc cot θ + sec θ tan θ, 0 < θ < π/2 dθ sin θ 8 cos θ −8 cos3 θ + sin3 θ + ; =− = 2 2θ cos sin θ sin2 θ cos2 θ dL = 0 if sin3 θ = 8 cos3 θ, tan3 θ = 8, tan θ = 2 which gives the dθ absolute minimum for L because lim L = lim − L = +∞. If θ→0+ √θ→π/2 √ tan θ = 2, then csc θ = √ 5/2 and sec θ = 5 so √ √ L = 8( 5/2) + 5 = 5 5 ft. 42. I=k cos φ 40. 2 L2 L L1 8 θ 1 Let x = number of steers per acre w = average market weight per steer T = total market weight per acre then T = xw where w = 2000 − 50(x − 20) = 3000 − 50x so T = x(3000 − 50x) = 3000x − 50x2 for 0 ≤ x ≤ 60, dT /dx = 3000 − 100x and dT /dx = 0 when x = 30. If x = 0, 30, 60 then T = 0, 45000, 0 so the total market weight per acre is largest when 30 steers per acre are allowed. 43. The daily proﬁt is P = (revenue) − (production cost) = 100x − (100, 000 + 50x + 0.0025x2 ) = −100, 000 + 50x − 0.0025x2 for 0 ≤ x ≤ 7000, so dP/dx = 50 − 0.005x and dP/dx = 0 when x = 10, 000. Because 10,000 is not in the interval [0, 7000], the maximum proﬁt must occur at an endpoint. When x = 0, P = −100, 000; when x = 7000, P = 127, 500 so 7000 units should be manufactured and sold daily. (b) Yes, because dP/dx > 0 when x = 7000 so proﬁt is increasing at this production level. (a) R(x) = px but p = 1000 − x so R(x) = (1000 −...
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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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