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Math 220 October 18 I. Optimization 1. A box with a square base and a top must have a volume of 27,000 cm3 . Find the dimensions of the box that minimize the amount of material used. 2. The sum of two positive numbers is 16. What is the smallest possible value of the sum their squares? 3. What is the maximum vertical distance between the line y = x and y = x3 for 0 ≤ x ≤ 1 4. Find the point on the x -axis that minimizes the sum of the distances to the points (0,1) and (2,0) 5. A farmer wishes to build a rectangular fence that encloses 100 ft2 , that has two parallel walls inside the area to divide the enclosed area into 3 equal areas. How should the fence be built to minimize the cost of the fence. II. Find the limit 1. x2 − 3x x→3 x − 3 lim 2. x4 − 3x3 + x2 + 1 x→1 x4 − 1 lim 3. lim x→0 1 sin(3x) tan(4x) 4. √ ln( x) lim x→∞ x2 5. t9 − 1 t→1 t3 − 1 lim 6. 6t − t2 t→0 t lim 7. 2u + sin(u) u→0 u + tan(u) lim 8. lim x ln(2x) x→0+ 9. lim x→0+ 10. 1 1 − x x e −1 a bx lim 1 + x→∞ x 11. lim (1 + ax)1/x x→∞ 2 I. Optimization 1. A box with a square base and a top must have a volume of 27,000 cm3 . Find the dimensions of the box that minimize the amount of material used. Answer: Let x be the length of side of a base and y be the length of a side. We have x2 y = 27000 The amount of material needed is 2x2 + 4xy M = 2x2 + 4xy 27000 M (x) = 2x2 + 4x 2 x 108000 2 M (x) = 2x + x 108000 0 M (x) = 4x − x2 108000 0 = 4x − x2 108000 4x = x2 3 x = 27000 x = 30 M ”(30) = 4 + 216000/303 > 0 So there is a minimum at x = 30. The dimensions are 30 by 30 by 30. 2. The sum of two postive numbers is 16. What is the smallest possible value of the sum their squares? Answers: We have x + y = 16 and we want to minimizes S = x2 + y 2 . 3 S = x2 + y 2 S = x2 + (16 − x)2 S 0 = 2x − 32 + 2x S 0 = 4x − 32 0 = 4x − 32 x=8 S 00 (8) = 4, so S has a min at x = 8. The smallest possible sum is S = 82 + 82 = 128 3. What is the maximum vertical distance between the line y = x and y = x3 for 0 ≤ x ≤ 1 Answers: The vertical distance between y = x and y = x3 is the distance between their y values: D = x − x3 D0 = 1 − 3x2 0 = 1 − 3x2 √ x = ±1/ 3 √ √ 00 00 D00 = √ D (1/ 3) < 0. There is a min at √ −6x, D (−1/ 3) > 0 and −1/ √3 and there is a max at 1/ 3. D(1/ 3) = 3−1/2 + 3−3/2 is the maximum distance. 4. Find the point on the x -axis that minimizes the sum of the distances to the points (0,1) and (2,0) Answer: 4 The sum of distances between (x, y) is D = √ x2 + 1+2−x, where x ≤ x √ x2 + 1 + 2 − x x D0 = √ −1 2 x +1 x 0= √ −1 2 x +1 D= x √ =1 2 x +1 √ x = x2 + 1 x2 = x 2 + 1 no solutions So D has no critical points. Notice D0 (x) < 0 when x < 2. So D is decreasing for x < 2. So we put the point at (2,0). 5. A farmer wishes to build a rectangular fence that encloses 100 ft2 , that has two parallel walls inside the area to divide the enclosed area into 3 equal areas. How should the fence be built to minimize the cost of the fence. Answers: Let x be the length of the parallel lines and y be the other side of the fence. So we have xy = 100 and to minimize the cost we must minimize the perimeter. P = 4x + 2y 5 P = 4x + 2y 100 P (x) = 4x + 2 x 200 0 P (x) = 4 − 2 x 200 0=4− 2 x 200 4= 2 x 4x2 = 200 x2 = 50 √ x = ± 50 √ P 00 (5 2) = 400 √ (5 2)3 √ > 0. So P has a minimum at x = 5 2 √ The fence should be built with the √ 4 parallel sides having length 5 2 and the other side having length 10 2 . II. Find the limit 1. x2 − 3x lim x→3 x − 3 Answer: x2 − 3x lim x→3 x − 3 2. 0 L 2x − 3 = lim x→3 0 1 =3 x4 − 3x3 + x2 + 1 x→1 x4 − 1 lim 6 Answer: x4 − 3x3 + x2 + 1 lim x→1 x4 − 1 3. 0 L 4x3 − 6x2 + 2x = lim x→1 0 4x3 0 = 4 =0 sin(3x) x→0 tan(4x) lim Answer: sin(3x) lim x→0 tan(4x) 4. cos(3x)3 0 L = lim x→0 sec2 (4x)4 0 3 = 4 √ ln( x) lim x→∞ x2 Answer: √ ln( x) ln(x) ∞ lim = lim x→∞ x→∞ 2x2 x2 ∞ 1/x L = lim x→∞ 4x 1 = lim x→∞ 4x2 =0 7 5. t9 − 1 lim 3 t→1 t − 1 Answer: t9 − 1 lim 3 t→1 t − 1 0 L 9t8 = lim 2 t→1 3t 0 = lim 3t6 t→1 =3 6. 6t − t2 t→0 t lim Answer: 6t − t2 1 lim t→0 t 0 t 2 6 −t =∞ lim+ t→0 t 6t − t2 lim− = −∞ t→0 t The limit does not exist. 7. 2u + sin(u) u→0 u + tan(u) lim Answer: 2u + sin(u) lim u→0 u + tan(u) 0 L 2 + cos(u) = lim u→0 1 + sec2 (u) 0 2+1 = 1+1 3 = 2 8 8. lim x ln(2x) x→0+ Answer: ln(2x) lim+ x ln(2x) (0(−∞)) = lim+ x→0 x→0 1/x 1/x L = lim+ x→0 −1/x2 −x2 = lim+ x→0 x = lim+ −x (−∞) ∞ x→0 =0 9. lim x→0+ 1 1 − x x e −1 Answer: lim x→0+ 10. 1 1 − x x e −1 ex − 1 − x 0 (∞ − ∞) = lim+ x x→0 xe − x 0 x e −1 L = lim+ x x→0 e + xex − 1 ex L = lim+ x→0 2ex + xex 1 = 2 a bx lim 1 + x→∞ x 9 Answer: a bx y = lim 1 + x→∞ x a bx ln(y) = ln( lim 1 + ) x→∞ x a bx = lim ln( 1 + ) x→∞ x a = lim bx ln(1 + ) x→∞ x ln(1 + xa ) = lim x→∞ 1/(bx) L = lim x→∞ = −a/x2 a 1/(1+ x ) −1/bx2 a −a(1+ x ) 2 x lim x→∞ −1/bx2 abx2 (1 + xa ) x→∞ x2 a = lim ab(1 + ) x→∞ x = ab = lim y = eln(y) = eab 11. lim (1 + ax)1/x x→∞ Answer: 10 y = lim (1 + ax)1/x x→∞ ln(y) = ln( lim (1 + ax)1/x ) x→∞ = lim ln((1 + ax)1/x ) x→∞ 1 ln (1 + ax) x ln (1 + ax) ∞ = lim x→∞ x ∞ = lim x→∞ L = lim a 1+ax 1 a = lim x→∞ 1 + ax =0 x→∞ y = eln(y) = e0 =1 11