Download ùñ dP 2 ¡ 2A S x 1 ùñ dS 1 ¡ 1 c r2a2 ¡a4 ùñ dA

Fall 2011, Math 1A - Calculus
GSI: Shishir Agrawal
Discussions 102 and 103
Worksheet 10
Problem 1. Find the dimensions of the rectangle of area 100 m2 whose perimeter is as small as possible.
Solution. See solution to problem 2.
Problem 2. Generalize your result in problem 1 to find the rectangle of arbitrary fixed area A whose
perimeter is as small as possible.
Solution. Let a and b be the lengths of the sides of the rectangle, so that the perimeter P is 2a
the area is ab A. This means that b A{a, so
P
2a
2
A
a
2a
200
a
ùñ
dP
da
2 2A
.
a2
2b and
?
Thus, the perimeter reaches a critical point when 2A{a2 2, which happens when a2 A, or?a A (we
exclude the negative square
? root since lengths must be positive). Observe that when?a A, dP {da is
negative, and when a ¡ A, dP {da is positive. Thus, by the first derivative test, a A corresponds to a
minimum perimeter.
?
?
?
Also, we know that b A{a A{ A A. Thus, the rectangle with a b A minimizes the
perimeter. Thus, for a given fixed area, a square is the rectangle which minimizes perimeter.
Problem 3. Find the positive number for which the sum of the number and its reciprocal is as small as
possible.
Solution. For a number x, we have
S
x
1
x
ùñ
dS
dx
1 x12 .
Thus, S reaches a critical point when 1{x2 1, which happens when x 1 (we exclude x 1 since we are
only looking for positive numbers). We could again argue that x 1 is a minimum by the first derivative
test, or, alternatively, we can observe that
d2 S
2
2
dx
x3
is always positive for x positive, so x 1 must be a minimum by the second derivative test. Thus, 1 is the
number we are looking for.
Problem 4. Find the dimensions of the rectangle inscribed in a semicircle of radius r whose area is as large
as possible. You may assume that one of the sides of the rectangle lies on the straight-edge of the semicircle.
Solution. Let a be the length of the side of the rectangle which is perpendicular to the straight-edge of the
semicircle, and b be the length?of the side which?lies along the straight-edge. Draw a couple of radii of the
circle and observe that b{2 r2 a2 , so b 2 r2 a2 . Thus, the area of the rectangle is given by
a
A ab 2a r2 a2
2
a
r2 a2 a4
ùñ
dA
da
p2r2 a 4a3 q 4a?
pr2 2a2 q 4?
pr2 2a2 q .
2?
r 2 a2 a4
a r 2 a2
r 2 a2
1
?
Thus, A has a critical point when r2 2a2 0, or when a2 r2 {2, which is equivalent to a r{ 2. Again,
verify that this is indeed a maximum (using either the first ?
or second derivative test), and then we see that
the dimensions of the rectangle maximizing area are a r{ 2 and
b
?
b 2 r2 pr{ 2q2
a
2
?
a
r2 r2 {2 2 r2 {2 r 2.
Problem 5. A piece of wire 10 m long is to be cut into two pieces. One piece is bent into a square, and the
other into a circle. How can the wire be cut so that the area enclosed by the two shapes is (a) minimized,
and (b) maximized?
Solution. Suppose the piece of wire is cut into pieces of length a and b, so that a b 10, and that the
piece of length a is bent into a square and the piece of length b is bent into a circle. This means that the
square has perimeter a, so has side length a{4, so has area a2 {16. The circle has circumference b, so has
radius r b{p2π q, so has area
b2
b2
.
π 2 4π
4π
Thus, the total area enclosed is
b2
a2
.
A
16 4π
Using the fact that a b 10, we can rewrite a with a 10 b, so then we have
A
p10 bq2
b2
.
4π
16
We are looking for the absolute minimum and maximum of A on the interval b P r0, 10s. To find critical
points, observe that A is just a quadratic polynomial in b, so it is differentiable everwhere, so it suffices to
find b where dA{db 0. We compute dA{db by
dA
db
bq
2p10
16
2b
4π
p108 bq
8b
2πb p8
b
.
2π
This equals zero when
10 b
8
2πb ðñ
20π 2πb 8b
Observe that
d2 A
db2
ðñ
dbd
20π
b 10
8
b
2π
2π qb
18
ðñ b 8 20π2π 410ππ .
1
2π
is always positive, so the graph of A is concave up; that is, it is an upwards facing parabola with a minimum
at b 10π {p4 π q. Thus, the minimum area enclosed is achieved when
b
10π
10π
and a 10 .
4 π
4 π
To find the maximum, we have to evaluate A at the endpoints b 0 and b 10. We have
A|b0
and clearly 16 ¡ 2π, so A|b10
2
2
10
100
1016 100
and A|b10 16
2π
2π
¡ A|b0 , so the maximum area enclosed is achieved when b 10 and a 0.
Problem 6. Find the equation of the line through the point p1, 1q which cuts off the least area from the
first quadrant.
2
Solution. Let p0, Y q be the y-intercept of the line and pX, 0q be the x-intercept.
p1, 1q and p0, Y q, so we can calculate the equation of the line to be
y p1 Y qx Y.
Thus, since this line also passes through pX, 0q, we have
Y Y .
0 p1 Y qX Y ðñ X 1Y
Y 1
The line passes through
Now, the area of cut off by the line can be computed by the formula for the area of a triangle by
A
1
XY
2
1
2
Y
Y
1
2
Y
2pYY 1q .
1 and when
4Y 2 4Y 2Y 2
pY 2q 0
dA
2Y p2pY 1qq Y 2 p2q
2Y
dY
4pY 1q2
4pY 1q2
4pY 1q2
which happens when Y 0 or Y 2. Thinking geometrically, we can exclude the critical points Y 0 and
Y 1 (both of these correspond to cases when the area in the first quadrant cut out by the line is actually
infinite). Observe that at Y 2, dA{dY changes from negative to positive (by inspecting the formula for
dA{dY above), so by the first derivative test, A achieves a minimum at Y 2. Thus, the equation of the
This has critical points at Y
line cutting off the least area from the first quadrant is
y
p1 2qx
2 x
2.
Problem 7. Find the dimensions of the cylinder inscribed inside a cone of height H and radius R whose
volume is as large as possible.
Solution. Suppose the cylinder has radius r and height h. Setting up similar triangles, we see that
r
R
so we can write
r
H H h
RpHH hq .
Then the volume of the cylinder is given by
V
πr h π
2
RpH hq
H
2
h
πR2
pH
H2
hq2 h.
We now compute using the product and chain rule
dV
dh
2
2
πR
p2pH hqh pH hq2 q πR
p2hpH hq pH hq2 q.
H2
H2
This equals zero if and only if 2hpH hq pH hq2 , and since H h, this happens if and only if 2h H h,
so h H {3. Thus,
RpH H {3q
{3 2R .
r
2RH
H
H
3
3