# Math 220 October 18 I. Optimization 1. A box with a square base and

```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.
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 &gt; 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?
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
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 = &plusmn;1/ 3
√
√
00
00
D00 =
√ D (1/ 3) &lt; 0. There is a min at
√ −6x, D (−1/ 3) &gt; 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)
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) &lt; 0 when x &lt; 2. So D is
decreasing for x &lt; 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.
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 = &plusmn; 50
√
P 00 (5 2) =
400
√
(5 2)3
√
&gt; 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
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
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
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
√
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
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
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
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+
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
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
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→∞
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
```