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# Assignment 8 Solutions.pdf

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School
Department
Calculus
Course
Calculus 1000A/B
Professor
Chris Brandl
Semester
Fall

Description
MATH 127 Fall 2012 Assignment 8 Solutions 1. Evaluate the following limits. ▯ ▯ 1 (a) lim xsin x!1 x This is of the type 1▯0 so we have an indeterminate product. We will manipulate ▯ ▯ 1 xsin so that we can apply L’Hopital’s rule. x ▯ ▯ ▯ 1 ▯ ▯ 1 sin x 0 x!1 xsin = x!1 1 = x x 0 ▯1▯▯▯1 ▯ cos x x = x!1 ▯1 ▯ ▯2 1 = lim cos x!1 x = cos(0) = 1 2 x (b) lim x x!1 ln(x) + e 1 This is a limit of the type so we use L’Hopital’s rule. 1 x2 2x ▯ 1 ▯ lim = lim = so we apply L Hopital s rule again x!1 ln(x) + ex x!1 x + ex 1 2 = lx!1 ▯1 x x2 + e = 0 3 (c) lim(x + e )x x!0+ 1 This is of the type 1so we have an indeterminate power. Notice that ▯ x 3▯ lim(x + e )x= lim eln (x+e ) x!0+ x!0 + xln(x+e ) = lim + x!0 1 3ln(x + e ) 0 Consider lim . This of the type so we apply L’Hopital’s rule. x!0+ x 0 x 1+e 3ln(x + e ) 3 x+ex lim = lim x!0 + x x!0+ 1 3(1 + e ) = lim x!0+ x + ex = 6 Thus, ▯ 3▯ x 3 ln (x+e ) 6 lim+(x + e ) = lim e+ = e x!0 x!0 2. A closed box is to be constructed so that its base length is three times the base width 3 and is to have a volume of 450 cm . Find the dimensions of the box so that it has the smallest possible surface area. Let w be the width, h be the height, and l be the length of the box. We then have that the length of the box is l = 3w. So then the volume is given by (3w)wh = 450 and the surface area is given by s = 2(3w ) + 2(3w)h + 2wh = 6w + 8wh. From the 150 volume equation we get h = w2 . Substituting this into the surface area equation we get 2 1200 s = 6w + w Di▯erentiating we get 3 3 0 1200 12w ▯ 1200 12(w ▯ 100) s = 12w ▯ 2 = 3 = 3 w w w p p 3 3 From this we see that the critical points are w = 0; 100 but w = 100 is the only 3 0 p3 reasonable value. We have that if 0 < w < 100 then s < 0 and if w > 100 then p p s > 0. From the ▯rst derivative test we see that w = 3100, l = 3 100, and h = 150 3 100 give the dimensions of the box that has the smallest surface area. 3. A rectangular sheet of metal with a perimeter of 36 cm and dimensions x and y is to be rolled into a cylinder. What dimensions of the sheet of metal would maximize the volume of the cylinder? 2 x y y We want to maximize v = ▯r y subject to the constraint that 2x + 2y = 36. Isolating for y in the perimeter equation we get y = 18 ▯ x. Substituting this into the volume equation we get v = ▯r (18 ▯ x). We want an equation for the volume in terms of a single variable. Notice that the circumference of the cylinder is given by 2▯r = x x which gives r = 2▯. Substituting this into the volume equation we get ▯ x ▯ 2 18x ▯ x 3 v = ▯ (18 ▯ x) = 2▯ 4▯ Di▯erentiating we get 2 v = 36x ▯ 3x = x(36 ▯ 3x) 4▯ 4▯ We see that the critical points are x = 0;12 but since x is the length of the rectangular sheet we get that x = 12 is the only reasonable value. We have that if 0 < x < 12 then v > 0 and if x > 12 then v < 0. Thus by the ▯rst derivative test we get that x = 12 maximizes the volume. Hence, the dimensions should be x = 12 and y = 6. 4. For the following functions (i) Find the domain, intercepts, and asymptotes of the function. (ii) Find the critical points, intervals where f is increasing and decreasing, and the local extrema. (iii) Find the intervals of concavity and in ection points of f. (iv) Sketch the graph. (a) f(x) = 2xe ▯x2 x ▯ 3x ▯ 4 (b) f(x) = 3x 3 ▯x2 (a) (i) Domain: f(x) = 2xe is de▯ned everywhere. Intercepts: f(0) = 2▯0e = 0, f(x) = 0 ) 2xe= 0 ) x = 0 f(x) passes through (0;0). Asymptotes: There are no vertical asymptotes. To ▯nd our horizontal asymptotes 2x 1 lim 2= x!1 ex 1 2 (L Hopital) = lim 2 = 0 x!1 2xex 2x Similarly, lim 2 = 0. x!▯1 ex Thus, y = 0 is a horizontal asymptote. (ii) Taking the derivative we get f (x) = 2ex2+ 2x(▯2x)e ▯x2= 2e▯x2▯ 4x e ▯x = e▯x2(2 ▯ 4x ) Looking for our critical points 0 2 p1 f (x) = 0 ) 2 ▯ 4x = 0 ) x = ▯ 2 ▯x2 2 0 e 2 ▯ 4x f (x) ▯1 x < p2 + ▯ ▯ p1 < x < p1 + + + 2 2 p x > 2 + ▯ ▯ ▯1
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