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2017
P1 Q3
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Maxima
17 P1 Q3
2017 H2 Mathematics Paper 1 Question 3
Differentiation II: Maxima, Minima, Rates of Change
Answers
(i)
x
=
1
2
,
{x=\frac{1}{\sqrt{2}}, \;}
x
=
2
1
,
x
=
−
1
2
{x=-\frac{1}{\sqrt{2}}}
x
=
−
2
1
(ii)
Maximum
Full solutions
(i)
y
2
−
2
x
y
+
5
x
2
−
10
=
0
\begin{equation} \qquad y^2 - 2xy + 5x^2 -10 = 0 \end{equation}
y
2
−
2
x
y
+
5
x
2
−
10
=
0
Differentiating w.r.t.
x
,
{x,}
x
,
2
y
d
y
d
x
−
2
x
d
y
d
x
−
2
y
+
10
x
=
0
2y\frac{\mathrm{d}y}{\mathrm{d}x} - 2x\frac{\mathrm{d}y}{\mathrm{d}x} - 2y + 10x = 0
2
y
d
x
d
y
−
2
x
d
x
d
y
−
2
y
+
10
x
=
0
At stationary points,
d
y
d
x
=
0
{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=0}
d
x
d
y
=
0
0
−
0
−
2
y
+
10
x
=
0
0-0-2y+10x=0
0
−
0
−
2
y
+
10
x
=
0
y
=
5
x
\begin{equation} y = 5x \end{equation}
y
=
5
x
Substituting
(
2
)
{(2)}
(
2
)
into
(
1
)
,
{(1),}
(
1
)
,
(
5
x
)
2
−
2
x
(
5
x
)
+
5
x
2
−
10
=
0
25
x
2
−
10
x
2
+
5
x
2
−
10
=
0
x
2
=
1
2
\begin{gather*} (5x)^2 - 2x(5x) + 5x^2 - 10 = 0 \\ 25x^2 - 10x^2 + 5x^2 - 10 = 0 \\ x^2 = \frac{1}{2} \\ \end{gather*}
(
5
x
)
2
−
2
x
(
5
x
)
+
5
x
2
−
10
=
0
25
x
2
−
10
x
2
+
5
x
2
−
10
=
0
x
2
=
2
1
x
=
1
2
■
or
x
=
−
1
2
■
\begin{align*} x &= \frac{1}{\sqrt{2}} \; \blacksquare \quad \textrm{or} \\ x &= -\frac{1}{\sqrt{2}} \; \blacksquare \end{align*}
x
x
=
2
1
■
or
=
−
2
1
■
(ii)
(
y
−
x
)
d
y
d
x
−
y
+
5
x
=
0
(
y
−
x
)
d
2
y
d
x
2
+
(
d
y
d
x
−
1
)
d
y
d
x
−
d
y
d
x
+
5
=
0
\begin{gather*} (y-x)\frac{\mathrm{d}y}{\mathrm{d}x} - y + 5x = 0 \\ (y-x)\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + \left( \frac{\mathrm{d}y}{\mathrm{d}x} - 1 \right)\frac{\mathrm{d}y}{\mathrm{d}x} - \frac{\mathrm{d}y}{\mathrm{d}x} + 5 = 0 \\ \end{gather*}
(
y
−
x
)
d
x
d
y
−
y
+
5
x
=
0
(
y
−
x
)
d
x
2
d
2
y
+
(
d
x
d
y
−
1
)
d
x
d
y
−
d
x
d
y
+
5
=
0
When
x
=
1
2
,
{x=\frac{1}{\sqrt{2}}, \;}
x
=
2
1
,
y
=
5
2
{y=\frac{5}{\sqrt{2}}}
y
=
2
5
and
d
y
d
x
=
0
{\frac{\mathrm{d}y}{\mathrm{d}x}=0}
d
x
d
y
=
0
(
5
2
−
1
2
)
d
2
y
d
x
2
+
0
−
0
+
5
=
0
(
4
2
)
d
2
y
d
x
2
=
−
5
\begin{gather*} \left( \frac{5}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + 0 - 0 + 5 = 0 \\ \left( \frac{4}{\sqrt{2}} \right) \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = -5 \\ \end{gather*}
(
2
5
−
2
1
)
d
x
2
d
2
y
+
0
−
0
+
5
=
0
(
2
4
)
d
x
2
d
2
y
=
−
5
d
2
y
d
x
2
=
−
5
2
4
<
0
\begin{align*} \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} &= -\frac{5\sqrt{2}}{4} \\ &< 0 \end{align*}
d
x
2
d
2
y
=
−
4
5
2
<
0
Hence the stationary point with
x
>
0
{x>0}
x
>
0
is a maximum
■
{\blacksquare}
■
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