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2011
P2 Q2
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Maxima
11 P2 Q2
2011 H2 Mathematics Paper 2 Question 2
Differentiation II: Maxima, Minima, Rates of Change
Answers
(ii)
x
=
n
2
−
3
6
{x = \frac{n}{2} - \frac{\sqrt{3}}{6}}
x
=
2
n
−
6
3
Full solutions
(i)
V
=
(
2
n
−
2
x
)
(
n
−
2
x
)
x
=
(
2
n
2
−
4
n
x
−
2
n
x
+
4
x
2
)
x
=
2
n
2
x
−
6
n
x
2
+
4
x
3
■
\begin{align*} V &= (2n-2x)(n-2x)x \\ &= (2n^2 - 4nx - 2nx + 4x^2)x \\ &= 2n^2x - 6nx^2 + 4x^3 \; \blacksquare \end{align*}
V
=
(
2
n
−
2
x
)
(
n
−
2
x
)
x
=
(
2
n
2
−
4
n
x
−
2
n
x
+
4
x
2
)
x
=
2
n
2
x
−
6
n
x
2
+
4
x
3
■
(ii)
d
V
d
x
=
2
n
2
−
12
n
x
+
12
x
2
\frac{\mathrm{d}V}{\mathrm{d}x} = 2n^2 - 12nx + 12x^2
d
x
d
V
=
2
n
2
−
12
n
x
+
12
x
2
At stationary values of
V
,
{V, }
V
,
d
V
d
x
=
0
{\frac{\mathrm{d}V}{\mathrm{d}x} = 0}
d
x
d
V
=
0
2
n
2
−
12
n
x
+
12
x
2
=
0
6
x
2
−
6
n
x
+
n
2
=
0
\begin{align*} 2n^2 - 12nx + 12x^2 &= 0 \\ 6x^2 - 6nx + n^2 &= 0 \end{align*}
2
n
2
−
12
n
x
+
12
x
2
6
x
2
−
6
n
x
+
n
2
=
0
=
0
x
=
6
n
±
36
n
2
−
4
(
6
)
(
n
2
)
12
=
6
n
±
12
n
2
12
=
6
n
±
2
n
3
12
=
n
2
±
3
6
\begin{align*} x &= \frac{6n \pm \sqrt{36n^2-4(6)(n^2)}}{12} \\ &= \frac{6n \pm \sqrt{12n^2}}{12} \\ &= \frac{6n\pm 2n\sqrt{3}}{12} \\ &= \frac{n}{2} \pm \frac{\sqrt{3}}{6} \end{align*}
x
=
12
6
n
±
36
n
2
−
4
(
6
)
(
n
2
)
=
12
6
n
±
12
n
2
=
12
6
n
±
2
n
3
=
2
n
±
6
3
If
x
=
n
2
+
3
6
,
{x= \frac{n}{2} + \frac{\sqrt{3}}{6},}
x
=
2
n
+
6
3
,
then the length cut out
(
2
x
)
{(2x)}
(
2
x
)
will be more than
n
{n}
n
which is impossible
Hence there is only one answer,
x
=
n
2
−
3
6
■
x = \frac{n}{2} - \frac{\sqrt{3}}{6} \; \blacksquare
x
=
2
n
−
6
3
■
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