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2015
P1 Q4
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Maxima
15 P1 Q4
2015 H2 Mathematics Paper 1 Question 4
Differentiation II: Maxima, Minima, Rates of Change
Answers
A
=
1
32
d
2
{A = \frac{1}{32} d^2}
A
=
32
1
d
2
Full solutions
Considering the perimeters of the rectangle and semicircle,
2
x
+
2
y
+
2
x
+
π
x
=
d
y
=
d
−
4
x
−
π
x
2
\begin{gather*} 2x + 2y + 2x + \pi x = d \\ y = \frac{d-4x-\pi x}{2} \end{gather*}
2
x
+
2
y
+
2
x
+
π
x
=
d
y
=
2
d
−
4
x
−
π
x
Let
A
{A}
A
denote the total area of the rectangle and semicircle
A
=
x
y
+
1
2
π
x
2
=
x
(
d
−
4
x
−
π
x
2
)
+
1
2
π
x
2
=
1
2
d
x
−
2
x
2
\begin{align*} A &= xy + \frac{1}{2} \pi x^2 \\ &= x \left(\frac{d-4x-\pi x}{2} \right) + \frac{1}{2} \pi x^2 \\ &= \frac{1}{2}dx - 2x^2 \end{align*}
A
=
x
y
+
2
1
π
x
2
=
x
(
2
d
−
4
x
−
π
x
)
+
2
1
π
x
2
=
2
1
d
x
−
2
x
2
d
A
d
x
=
1
2
d
−
4
x
\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{1}{2}d - 4x
d
x
d
A
=
2
1
d
−
4
x
At stationary values of
A
,
{A, }
A
,
d
A
d
x
=
0
{\displaystyle \frac{\mathrm{d}A}{\mathrm{d}x} = 0}
d
x
d
A
=
0
1
2
d
−
4
x
=
0
x
=
1
8
d
\begin{gather*} \frac{1}{2}d - 4x = 0 \\ x = \frac{1}{8}d \end{gather*}
2
1
d
−
4
x
=
0
x
=
8
1
d
A
=
1
2
d
(
1
8
d
)
−
2
(
1
8
d
)
2
=
1
32
d
2
■
\begin{align*} A &= \frac{1}{2}d\left(\frac{1}{8}d\right) - 2 \left(\frac{1}{8}d\right)^2 \\ &= \frac{1}{32} d^2 \; \blacksquare \end{align*}
A
=
2
1
d
(
8
1
d
)
−
2
(
8
1
d
)
2
=
32
1
d
2
■
d
2
A
d
x
2
=
−
4
<
0
\begin{align*} \frac{\mathrm{d}^{2}A}{\mathrm{d}x^{2}} &= -4 \\ <0 \end{align*}
d
x
2
d
2
A
<
0
=
−
4
Hence
A
=
1
32
d
2
{A= \frac{1}{32} d^2}
A
=
32
1
d
2
is a maximum
■
{\blacksquare}
■
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