A plane figure is bounded by the x-axis and the curve x^2+y=9. Consider a set of all rectangles inscribed into the figure so that a side of each rectangle lies on the x-axis and two of its vertices are on the curve x^2+y=9. By rotating these rectangles around the y axis, we get a set M of solids of revolution. Express the volume of a solid of revolution from M as a function V of a variable x, where 2x is the length of the side-lying on the x-axis.
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Let us set up the following variables:
{
P
(
x
,
y
)
coordinate of the right hand corner
A
Area of Rectangle
P
lies on the parabola and
y
=
12
−
x
2
, so
P
=
P
(
x
,
12
−
x
2
)
Due to symmetry The width of the rectangle is half the distance between P and the y-axis, ie
Width =
2
x
and Height=
y
Hence the Area of the rectangle is:
A
=
W
d
i
t
h
×
H
e
i
g
h
t
∴
A
=
2
x
y
∴
A
=
2
x
(
12
−
x
2
)
∴
A
=
24
x
−
2
x
3
)
..... [1]
We are asked to maximise the Area as
x
changes so hopefully we can identify a critical point of
A
associated with a maximum, So we need to find
d
A
d
x
Differentiating [1] wrt
x
∴
d
A
d
x
=
24
−
6
x
2
..... [2]
At a critical point,
d
A
d
x
=
0
∴
24
−
6
x
2
=
0
∴
6
x
2
=
24
∴
x
2
=
4
∴
x
=
±
2
Obviously
x
must be positive (otherwise we have an imaginary rectangle with negative area for a box that has collapsed in on itself)
∴
x
=
2
We need to check if this is a max or a min, so differentiate [2] wrt
x
to get;
∴
d
2
A
d
x
2
=
−
12
x
∴
d
2
A
d
x
2
=
−
12
x
<
0
when
x
=
2
, confirming a max
When
x
=
2
we have:
Width =
2
⋅
2
=
4
Height =
12
−
4
=
8
Area =
32
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