Math 220 March 7 I. If 1200 cm2 of material is available to make a ...

Math 220 March 7
I. If 1200 cm 2 of material is available to make a box with a square base
and an open top, find the largest possible volume of the box.
II. Find the point on the curve y = √ x that is closest to point (4,0).
III. At which points on the curve y = 1 + x + 60x 3 − x 5 does the tangent
line have the largest slope?
IV. A farmer whats to fence in area of 400 square feet in a rectangular
field and then divide it 3 sections with a two fence parallel to one of the sides
of the rectangle. How can he do this so as to minimize the cost of the fence?
V. A Frisbee has landed in the ocean 150 from the shore. A dog is standing
200 feet down the shore from where the Frisbee landed. The dog can run
5 feet per second and swim 2 feet per second. How far should the dog run
down the shore before he get jumps in the water to get the Frisbee in the
least amount of time?
VI. Find the limit.
1.
2.
3.
4.
5.
e
lim
x→∞
x
x3 sin(3x)
lim
x→0 cos(4x)
lim
x→0 +
ln( √ x)
x2 5
lim
x→0
x − 7x x
sin(x) − x
lim
x→0 x
1

6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
lim
x→0
sin(x
lim
x→0
2 )
x
lim
x→0 +
cos(x)
x
√ 1 + 2x − √ 1 − 4x
x
2x − sin(x)
lim
x→0 2x − tan(x)
lim
x→0 +
1 1
−
x ex
− 1
lim (sin(x) ln(x))
x→0
lim
x→0 +
2
x − ln(x )
lim (tan(3x))6x
x→0 +
lim (x)2/(1−x)
x→1 +
x 2
lim e + x
x→∞
1/(1−x) 2

I. If 1200 cm 2 of material is available to make a box with a square base
and an open top, find the largest possible volume of the box.
Answer:
V olume = x 2 y
SurfaceArea = 1200 = 4x 2 + xy =⇒ y =
v = x 2 y
v(x) = x 2 1200 − 4x2
( )
x
v(x) = x(1200 − 4x 2 )
v ′ (x) = (1200 − 4x 2 ) + x(−8x)
0 = 1200 − 12x 2
0 = 100 − x 2
x = ±10
v ′′ (x) = −24x
v ′′ (10) < 0
1200 − 4x2
x
There is a maximum when x = 10 and y = 12
The largest possible volume of the box is 1200 cm 3 .
II. Find the point on the curve y = √ x that is closest to point (4,0).
Answer:
Distance = (x − 4) 2 + (y − 0) 2
3

D = (x − 4) 2 + (y − 0) 2
D(x) = (x − 4) 2 + ( √ x) 2
D(x) = √ x 2 − 8x + 16 + x
D(x) = √ x 2 − 7x + 16
D ′ 2x − 7
(x) =
2 √ x2 − 7x + 16
2x − 7
0 =
2 √ x2 − 7x + 16
0 = 2x − 7
x = 7
2
D ′ (x) < 0 when x < 7/2
D ′ (x) > 0 when x > 7/2
Thus there is a minimum at x = 7/2
The point (7/2, 7/2) is closest.
III. At which points on the curve y = 1 + x + 60x 3 − x 5 does the tangent
line have the largest slope?
Answer:
slope = y ′ = 180x 2 − 5x 4
4

m ′′ (x) = 360 − 60x 2
m = 180x 2 − 5x 4
m ′ (x) = 360x − 20x 3
m ′ (x) = 20x(18 − x 2 )
m ′′ (0) > 0
m ′′ ( √ 18) < 0
m ′′ ( √ 18) < 0
0 = 20x(18 − x 2 )
x = 0, ± √ 18
There is a maximum at x = ± √ 18
So the tangent line will have the largest slope at the points
( √ 18, 1+ √ 18+60(18) 3/2 −(18) 5/2 ) and (− √ 18, 1+ √ 18+60(18) 3/2 −(18) 5/2 )
IV. A farmer whats to fence in area of 400 square feet in a rectangular
field and then divide it 3 sections with a two fence parallel to one of the sides
of the rectangle. How can he do this so as to minimize the cost of the fence?
Answer:
400 = xy
Cost = 4x + 2y
5

A ′′ (x) = 1600x −3 > 0
C = 4x + 2y
C(x) = 4x + 800
x
C ′ (x) = 4 − 800x −2
0 = 4 − 800x −2
x = 4x 2 − 800
200 = x 2
x = ± √ 200
There is maximum at x = √ 200
Cost is be minimize if the fence is built to be √ 200 by 2 √ 200 where √ 200 is
the length of the 4 parallel sections.
V. A Frisbee has landed in the ocean 150 from the shore. A dog is standing
200 feet down the shore from where the Frisbee landed. The dog can run
5 feet per second and swim 2 feet per second. How far should the dog run
down the shore before he get jumps in the water to get the Frisbee in the
least amount of time?
Answer:
T ime = 1 1
(x) + 1502 + (200 − x) 2
5 2
6

T (x) = 1
5
T ′ (x) = 1
5 +
1
(x) + 1502 + (200 − x) 2
2
−400 + 2x
4 1502 + (200 − x) 2
0 = 1
5 +
−400 + 2x
4 1502 + (200 − x) 2
1
5 =
400 − 2x
4 1502 + (200 − x) 2
( 1
5 )2 400 − 2x
= (
4 1502 )2
+ (200 − x) 2
1
25 =
4(200 − x) 2
16(1502 + (200 − x) 2 )
1
25 =
(200 − x) 2
4(1502 + (200 − x) 2 )
4(150 2 + (200 − x) 2 ) = 25(200 − x) 2
300 2 + 4(200 − x) 2 = 25(200 − x) 2
21(200 − x) 2 = 300 2
(200 − x) 2 = 30000
7
30000
200 − x = ±
7
30000
x = 200 ±
7
x = 134.535, 265.465
30000
T (x) < 0 when x < 200 −
7
30000
T (x) > 0 when x > 200 −
7
There is a minimum at x = 134.535
The dog should run 134.535 feet before jumping in the water.
7

VI. Find the limit.
1.
2.
3.
Answer:
Answer:
e
lim
x→∞
x
x3 L.H.
= lim
x→∞
L.H.
= lim
x→∞
L.H.
= lim
x→∞
= 1
6
e
lim
x→∞
x
x3
∞
∞
ex 3x2
∞
∞
ex
∞
6x ∞
ex 6
sin(3x)
lim
x→0 cos(4x)
sin(3x)
lim
x→0 cos(4x)
= 0
1
= 0
lim
x→0 +
ln( √ x)
x2 8

4.
5.
Answer:
Answer:
Answer:
lim
x→0 +
ln( √ x)
x2 = lim
x→0 +
(1/2) ln(x)
= lim
x→0 +
= lim
x→0 +
= −∞
5
lim
x→0
x − 7x x 2
ln(x)
2x2 1 1
(ln(x) − ) (∞ · −∞)
x 2x
5
lim
x→0
x − 7x x
x
L.H.
= lim
x→0
0
0
= ln(5) − ln(7)
ln(5)5 x − ln(7)7 x
1
sin(x) − x
lim
x→0 x
sin(x) − x 0
lim
x→0 x 0
L.H. cos(x) − 1
= lim
x→0 1
= 0
9

6.
7.
8.
Answer:
Answer:
Answer:
lim
x→0
lim
x→0
sin(x
lim
x→0
2 )
x
sin(x
lim
x→0
2 )
x
L.H.
= lim
x→0
= 0
0
0
2x cos(x2 )
1
lim
x→0 +
cos(x)
x
lim
x→0 +
cos(x) 1
x 0 +
= ∞
√ 1 + 2x − √ 1 − 4x
x
√ 1 + 2x − √ 1 − 4x
L.H.
= lim
=
2
2
x→0
+ 4
2
1
= 1 + 2
= 3
x
2
2 √ −4 − 1+2x 2 √ 1−4x
1
10
0
0

9.
10.
11.
Answer:
Answer:
2x − sin(x)
lim
x→0 2x − tan(x)
2x − sin(x)
lim
x→0 2x − tan(x)
= lim
x→0
= 1
1
= 1
0
0
2 − cos(x)
2 − sec2 (x)
lim
x→0 +
1 1
−
x ex
− 1
lim
x→0 +
1 1
−
x ex
− 1
= lim
x→0 +
x e − 1 − x
x(ex
0
− 1) 0
L.H.
= lim
x→0 +
ex − 1
ex − 1 + xex
0
0
L.H.
= lim
x→0 +
= 1
2
e x
e x + e x + xe x
lim (sin(x) ln(x))
x→0 +
11

12.
13.
Answer:
Answer:
Answer:
lim (sin(x) ln(x))
x→0 +
= lim
x→0 +
ln(x)
csc(x)
L.H.
= lim
x→0 +
1/x
− csc(x) cot(x)
= lim
x→0 +
− sin2
(x) 0
x cos(x) 0
L.H.
= lim
x→0 +
−2 sin(x) cos(x)
cos(x) − x sin(x)
= 0
1
= 0
lim
x→0 +
2
x − ln(x )
lim
x→0 +
2
x − ln(x )
= lim (x − 2 ln(x))
x→0 +
= 0 − (−∞)
= ∞
lim (tan(3x))6x
x→0 +
12

14.
Answer:
ln(y) = ln( lim
x→0 + (tan(3x))6x )
= lim 6x ln(tan(3x)) (0(−∞))
x→0 +
= lim
x→0 +
6 ln(tan(3x))
x −1
−∞
∞
L.H.
= lim
x→0 +
18 sec2 (3x)
tan(3x)
−x−2 = lim
x→0 +
−18x2 sec2 (3x)
tan(3x)
= lim
x→0 +
−18x2 cos(3x)
sin(3x) cos2 (3x)
= lim
x→0 +
−18x2
0
sin(3x) cos(3x) 0
L.H.
= lim
x→0 +
−36x
cos2 (3x) − sin2 (3x)
= 0
1
= 0
y = e ln(y)
= e 0
= 1
lim (x)2/(1−x)
x→1 +
13

15.
Answer:
ln(y) = ln( lim
x→1 + (x)2/(1−x) )
= lim
x→1 +
2
1 − x ln(x)
= lim
x→1 +
2 ln(x) 0
1 − x 0
L.H.
= lim
x→1 +
2/x
−1
= −2
y = e ln(y)
= e −2
x 2
lim e + x
x→∞
1/(1−x) 14

x 2
ln(y) = ln( lim e + x
x→∞
1/(1−x) )
= lim
x→∞
= lim
x→∞
1
1 − x ln e x + x 2
ln (ex + x2
) ∞
1 − x −∞
e
L.H.
= lim
x→∞
x +2x
ex +x2 −1
−e
= lim
x→∞
x − 2x
ex + x2
−∞
∞
L.H. −e
= lim
x→∞
x − 2
ex
−∞
+ 2x ∞
−ex ex
−∞
+ 2 ∞
L.H.
= lim
x→∞
L.H.
= lim
x→∞
= lim
x→∞ −1
= −1
y = e ln(y)
= e −1
15
−e x
e x