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2013
P2 Q2
Topical
Maxima
13 P2 Q2
2013 H2 Mathematics Paper 2 Question 2
Differentiation II: Maxima, Minima, Rates of Change
Answers
(ii)
Maximum
V
=
1
54
a
3
{\textrm{Maximum } V = \frac{1}{54} a^3}
Maximum
V
=
54
1
a
3
Full solutions
(i)
Let
y
{y}
y
denote the length that is cut
tan
3
0
∘
=
x
y
1
3
=
x
y
y
=
x
3
\begin{align*} \tan 30^\circ &= \frac{x}{y} \\ \frac{1}{\sqrt{3}} &= \frac{x}{y} \\ y &= x\sqrt{3} \end{align*}
tan
3
0
∘
3
1
y
=
y
x
=
y
x
=
x
3
V
=
1
2
(
a
−
2
y
)
2
sin
6
0
∘
x
=
1
2
(
a
−
2
x
3
)
2
3
2
x
=
1
4
x
3
(
a
−
2
x
3
)
2
■
\begin{align*} V &= \frac{1}{2} \left( a - 2y \right)^2 \sin 60^\circ \, x \\ &= \frac{1}{2} \left( a - 2x\sqrt{3} \right)^2 \frac{\sqrt{3}}{2} \, x\\ &= \frac{1}{4} x \sqrt{3} \left( a - 2x\sqrt{3} \right)^2 \; \blacksquare \end{align*}
V
=
2
1
(
a
−
2
y
)
2
sin
6
0
∘
x
=
2
1
(
a
−
2
x
3
)
2
2
3
x
=
4
1
x
3
(
a
−
2
x
3
)
2
■
(ii)
d
V
d
x
=
1
4
3
(
a
−
2
x
3
)
2
+
1
2
x
3
(
a
−
2
x
3
)
(
−
2
3
)
=
1
4
3
(
a
−
2
x
3
)
2
−
3
x
(
a
−
2
x
3
)
=
(
a
−
2
x
3
)
(
1
4
3
(
a
−
2
x
3
)
−
3
x
)
=
(
a
−
2
x
3
)
(
a
4
3
−
3
2
x
−
3
x
)
=
(
a
−
2
x
3
)
(
a
4
3
−
9
2
x
)
\begin{align*} \frac{\mathrm{d}V}{\mathrm{d}x} &= \frac{1}{4}\sqrt{3} \left( a - 2x\sqrt{3} \right)^2 + \frac{1}{2} x \sqrt{3} \left( a - 2x\sqrt{3} \right) \left( -2\sqrt{3} \right) \\ &= \frac{1}{4}\sqrt{3} \left( a - 2x\sqrt{3} \right)^2 - 3x \left( a-2x\sqrt{3} \right) \\ &= \left( a-2x\sqrt{3} \right) \left( \frac{1}{4}\sqrt{3} \left(a - 2x \sqrt{3}\right) - 3x \right) \\ &= \left( a-2x\sqrt{3} \right) \left( \frac{a}{4}\sqrt{3} - \frac{3}{2}x - 3x \right) \\ &= \left( a-2x\sqrt{3} \right) \left( \frac{a}{4}\sqrt{3} - \frac{9}{2} x \right) \\ \end{align*}
d
x
d
V
=
4
1
3
(
a
−
2
x
3
)
2
+
2
1
x
3
(
a
−
2
x
3
)
(
−
2
3
)
=
4
1
3
(
a
−
2
x
3
)
2
−
3
x
(
a
−
2
x
3
)
=
(
a
−
2
x
3
)
(
4
1
3
(
a
−
2
x
3
)
−
3
x
)
=
(
a
−
2
x
3
)
(
4
a
3
−
2
3
x
−
3
x
)
=
(
a
−
2
x
3
)
(
4
a
3
−
2
9
x
)
At stationary values of
V
,
{V, }
V
,
d
V
d
x
=
0
{\displaystyle \frac{\mathrm{d}V}{\mathrm{d}x} = 0}
d
x
d
V
=
0
(
a
−
2
x
3
)
(
a
4
3
−
9
2
x
)
=
0
\left( a-2x\sqrt{3} \right) \left( \frac{a}{4}\sqrt{3} - \frac{9}{2} x \right) = 0
(
a
−
2
x
3
)
(
4
a
3
−
2
9
x
)
=
0
Since
a
−
2
x
3
≠
0
{a - 2x \sqrt{3} \neq 0}
a
−
2
x
3
=
0
(otherwise
V
=
0
{V=0}
V
=
0
),
a
4
3
−
9
2
x
=
0
x
=
1
18
a
3
\begin{gather*} \frac{a}{4}\sqrt{3} - \frac{9}{2} x = 0 \\ x = \frac{1}{18} a \sqrt{3} \end{gather*}
4
a
3
−
2
9
x
=
0
x
=
18
1
a
3
V
=
1
4
x
3
(
a
−
2
x
3
)
2
=
1
4
(
1
18
a
3
)
3
(
a
−
2
(
1
18
a
3
)
3
)
2
=
1
24
a
(
a
−
1
3
a
)
2
=
1
24
a
(
2
3
a
)
2
=
1
54
a
3
■
\begin{align*} V &= \frac{1}{4} x \sqrt{3} \left( a - 2x\sqrt{3} \right)^2 \\ &= \frac{1}{4} \left(\frac{1}{18} a \sqrt{3}\right) \sqrt{3} \Bigg( a - 2\left(\frac{1}{18} a \sqrt{3}\right)\sqrt{3} \Bigg)^2 \\ &= \frac{1}{24} a \left( a - \frac{1}{3}a \right)^2 \\ &= \frac{1}{24} a \left( \frac{2}{3}a \right)^2 \\ &= \frac{1}{54} a^3 \; \blacksquare \end{align*}
V
=
4
1
x
3
(
a
−
2
x
3
)
2
=
4
1
(
18
1
a
3
)
3
(
a
−
2
(
18
1
a
3
)
3
)
2
=
24
1
a
(
a
−
3
1
a
)
2
=
24
1
a
(
3
2
a
)
2
=
54
1
a
3
■
d
2
V
d
x
2
=
−
2
3
(
a
4
3
−
9
2
x
)
−
9
2
(
a
−
2
x
3
)
=
−
3
2
a
+
9
x
3
−
9
2
a
+
9
x
3
=
−
6
a
+
18
x
3
\begin{align*} \frac{\mathrm{d}^{2}V}{\mathrm{d}x^{2}} &= -2\sqrt{3} \left( \frac{a}{4}\sqrt{3} - \frac{9}{2} x \right) - \frac{9}{2} \left( a-2x\sqrt{3} \right) \\ &= - \frac{3}{2} a + 9 x \sqrt{3} - \frac{9}{2} a + 9 x \sqrt{3} \\ &= - 6 a + 18x \sqrt{3} \end{align*}
d
x
2
d
2
V
=
−
2
3
(
4
a
3
−
2
9
x
)
−
2
9
(
a
−
2
x
3
)
=
−
2
3
a
+
9
x
3
−
2
9
a
+
9
x
3
=
−
6
a
+
18
x
3
When
x
=
1
18
a
3
,
{x= \frac{1}{18} a \sqrt{3},}
x
=
18
1
a
3
,
d
2
V
d
x
2
=
−
6
a
+
3
a
=
−
3
a
<
0
\begin{align*} \frac{\mathrm{d}^{2}V}{\mathrm{d}x^{2}} &= -6a + 3a \\ &= -3a \\ &< 0 \end{align*}
d
x
2
d
2
V
=
−
6
a
+
3
a
=
−
3
a
<
0
Hence
V
=
1
54
a
3
{V = \frac{1}{54} a^3}
V
=
54
1
a
3
is a maximum
■
{\blacksquare}
■
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