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2014
P1 Q11
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Maxima
14 P1 Q11
2014 H2 Mathematics Paper 1 Question 11
Differentiation II: Maxima, Minima, Rates of Change
Answers
(i)
V
=
1
3
π
r
2
16
−
r
2
+
2
3
π
r
3
{V}\allowbreak {= \frac{1}{3} \pi r^2 \sqrt{16-r^2}}\allowbreak {+ \frac{2}{3}\pi r^3 }
V
=
3
1
π
r
2
16
−
r
2
+
3
2
π
r
3
(ii)
r
1
=
1.207
,
{r_1 = 1.207, \;}
r
1
=
1.207
,
r
2
=
3.951
{r_2 = 3.951}
r
2
=
3.951
(iii)
r
1
=
3.951
,
{r_1 = 3.951, \;}
r
1
=
3.951
,
h
=
0.625
{h=0.625}
h
=
0.625
(iv)
Full solutions
(i)
h
=
4
2
−
r
2
=
16
−
r
2
\begin{align*} h &= \sqrt{4^2 - r^2} \\ &= \sqrt{16-r^2} \end{align*}
h
=
4
2
−
r
2
=
16
−
r
2
V
=
1
3
π
r
2
h
+
2
3
π
r
3
V
=
1
3
π
r
2
16
−
r
2
+
2
3
π
r
3
■
\begin{align*} V &= \frac{1}{3} \pi r^2 h + \frac{2}{3}\pi r^3 \\ V &= \frac{1}{3} \pi r^2 \sqrt{16-r^2} + \frac{2}{3}\pi r^3 \; \blacksquare \end{align*}
V
V
=
3
1
π
r
2
h
+
3
2
π
r
3
=
3
1
π
r
2
16
−
r
2
+
3
2
π
r
3
■
d
V
d
x
=
2
3
π
r
16
−
r
2
+
−
2
r
(
π
r
2
)
2
(
3
)
16
−
r
2
+
2
π
r
2
=
2
π
r
(
16
−
r
2
)
−
π
r
3
3
16
−
r
2
+
2
π
r
2
=
π
r
(
32
−
3
r
2
)
3
16
−
r
2
+
2
π
r
2
\begin{align*} \frac{\mathrm{d}V}{\mathrm{d}x} &= \frac{2}{3}\pi r \sqrt{16-r^2} + \frac{-2r (\pi r^2)}{2(3)\sqrt{16-r^2}} + 2\pi r^2 \\ &= \frac{2\pi r (16-r^2) - \pi r^3}{3\sqrt{16-r^2}} + 2\pi r^2 \\ &= \frac{\pi r (32-3r^2)}{3\sqrt{16-r^2}} + 2\pi r^2 \\ \end{align*}
d
x
d
V
=
3
2
π
r
16
−
r
2
+
2
(
3
)
16
−
r
2
−
2
r
(
π
r
2
)
+
2
π
r
2
=
3
16
−
r
2
2
π
r
(
16
−
r
2
)
−
π
r
3
+
2
π
r
2
=
3
16
−
r
2
π
r
(
32
−
3
r
2
)
+
2
π
r
2
At maximum,
d
V
d
x
=
0
,
{\displaystyle \frac{\mathrm{d}V}{\mathrm{d}x}=0,}
d
x
d
V
=
0
,
π
r
(
32
−
3
r
2
)
3
16
−
r
2
+
2
π
r
2
=
0
2
π
r
2
=
π
r
(
3
r
2
−
32
)
3
16
−
r
2
6
r
=
3
r
2
−
32
16
−
r
2
36
r
2
=
(
3
r
2
−
32
)
2
16
−
r
2
576
r
2
−
36
r
4
=
9
r
4
−
192
r
2
+
1024
\begin{gather*} \frac{\pi r (32-3r^2)}{3\sqrt{16-r^2}} + 2\pi r^2 = 0 \\ 2\pi r^2 = \frac{\pi r (3r^2-32)}{3\sqrt{16-r^2}} \\ 6 r = \frac{3r^2-32}{\sqrt{16-r^2}} \\ 36r^2 = \frac{(3r^2-32)^2}{16-r^2} \\ 576r^2 - 36r^4 = 9r^4 - 192r^2 + 1024 \\ \end{gather*}
3
16
−
r
2
π
r
(
32
−
3
r
2
)
+
2
π
r
2
=
0
2
π
r
2
=
3
16
−
r
2
π
r
(
3
r
2
−
32
)
6
r
=
16
−
r
2
3
r
2
−
32
36
r
2
=
16
−
r
2
(
3
r
2
−
32
)
2
576
r
2
−
36
r
4
=
9
r
4
−
192
r
2
+
1024
Hence
r
1
{r_1}
r
1
satisfies the equation
45
r
4
−
768
r
2
+
1024
=
0
■
45r^4 - 768r^2 + 1024 = 0 \; \blacksquare
45
r
4
−
768
r
2
+
1024
=
0
■
(ii)
Using a GC, for
r
>
0
,
{r>0,}
r
>
0
,
r
1
=
1.207
(3 dp)
■
or
r
1
=
3.951
(3 dp)
■
\begin{align*} r_1 &= 1.207 \textrm{ (3 dp)} \; \blacksquare \quad \textrm{or} \\ r_1 &= 3.951 \textrm{ (3 dp)} \; \blacksquare \quad \end{align*}
r
1
r
1
=
1.207
(3 dp)
■
or
=
3.951
(3 dp)
■
(iii)
When
r
1
=
1.207
{r_1 = 1.207}
r
1
=
1.207
d
V
d
x
=
π
1.207
(
32
−
3
(
1.207
)
2
)
3
16
−
(
1.207
)
2
+
2
π
(
1.207
)
2
=
18.320
≠
0
\begin{align*} \frac{\mathrm{d}V}{\mathrm{d}x} &= \frac{\pi 1.207 (32-3(1.207)^2)}{3\sqrt{16-(1.207)^2}} + 2\pi (1.207)^2 \\ &= 18.320 \\ &\neq 0 \end{align*}
d
x
d
V
=
3
16
−
(
1.207
)
2
π
1.207
(
32
−
3
(
1.207
)
2
)
+
2
π
(
1.207
)
2
=
18.320
=
0
Hence
r
1
=
1.207
{r_1 = 1.207}
r
1
=
1.207
does not give a stationary value of V
■
{\blacksquare}
■
Meanwhile, when
r
1
=
3.951
{r_1 = 3.951}
r
1
=
3.951
d
V
d
x
=
π
3.951
(
32
−
3
(
3.951
)
2
)
3
16
−
(
3.951
)
2
+
2
π
(
3.951
)
2
=
0.00349
≈
0
\begin{align*} \frac{\mathrm{d}V}{\mathrm{d}x} &= \frac{\pi 3.951 (32-3(3.951)^2)}{3\sqrt{16-(3.951)^2}} + 2\pi (3.951)^2 \\ &= 0.00349 \\ &\approx 0 \end{align*}
d
x
d
V
=
3
16
−
(
3.951
)
2
π
3.951
(
32
−
3
(
3.951
)
2
)
+
2
π
(
3.951
)
2
=
0.00349
≈
0
r
1
=
3.951
■
h
=
16
2
−
3.951
2
=
0.625
■
\begin{align*} r_1 &= 3.951 \; \blacksquare \\ h &= \sqrt{16^2 - 3.951^2} \\ &= 0.625 \; \blacksquare \end{align*}
r
1
h
=
3.951
■
=
1
6
2
−
3.95
1
2
=
0.625
■
(iv)
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