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2010
P1 Q9
Topical
Maxima
10 P1 Q9
2010 H2 Mathematics Paper 1 Question 9
Differentiation II: Maxima, Minima, Rates of Change
Answers
(i)
x
=
200
3
(
k
+
1
)
3
{x=\sqrt[3]{\frac{200}{3}(k+1)}}
x
=
3
3
200
(
k
+
1
)
(ii)
y
x
=
3
2
(
k
+
1
)
{\frac{y}{x} = \frac{3}{2(k+1)}}
x
y
=
2
(
k
+
1
)
3
(iii)
3
4
≤
y
x
<
3
2
{\frac{3}{4} \leq \frac{y}{x} < \frac{3}{2}}
4
3
≤
x
y
<
2
3
(iv)
k
=
1
2
{k=\frac{1}{2}}
k
=
2
1
Full solutions
(i)
Volume
=
300
3
x
2
y
=
300
\begin{align*} \textrm{Volume} &= 300 \\ 3x^2 y &= 300 \end{align*}
Volume
3
x
2
y
=
300
=
300
y
=
100
x
2
\begin{equation} y = \frac{100}{x^2} \end{equation}
y
=
x
2
100
Let
A
{A}
A
denote the total external surface area of the box and the lid
A
=
2
x
y
+
2
(
3
x
y
)
+
2
(
3
x
2
)
+
2
(
k
x
y
)
+
2
(
3
k
x
y
)
=
8
x
y
+
6
x
2
+
8
k
x
y
\begin{equation}\begin{split} \qquad A &= 2xy + 2(3xy) + 2(3x^2) + 2(kxy) + 2(3kxy) \\ &= 8xy + 6x^2 + 8kxy \end{split}\end{equation}
A
=
2
x
y
+
2
(
3
x
y
)
+
2
(
3
x
2
)
+
2
(
k
x
y
)
+
2
(
3
k
x
y
)
=
8
x
y
+
6
x
2
+
8
k
x
y
Substituting
(
1
)
{(1)}
(
1
)
into
(
2
)
,
{(2),}
(
2
)
,
A
=
8
x
(
100
x
2
)
+
6
x
2
+
8
k
x
(
100
x
2
)
=
800
(
k
+
1
)
x
+
6
x
2
\begin{align*} A &= 8x \left( \frac{100}{x^2} \right) + 6x^2 + 8kx\left( \frac{100}{x^2} \right) \\ &= \frac{800(k+1)}{x} + 6x^2 \end{align*}
A
=
8
x
(
x
2
100
)
+
6
x
2
+
8
k
x
(
x
2
100
)
=
x
800
(
k
+
1
)
+
6
x
2
d
A
d
x
=
−
800
(
k
+
1
)
x
2
+
12
x
\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{-800(k+1)}{x^2} + 12x
d
x
d
A
=
x
2
−
800
(
k
+
1
)
+
12
x
At minimum
A
,
d
A
d
x
=
0
,
{A, \frac{\mathrm{d}A}{\mathrm{d}x}=0,}
A
,
d
x
d
A
=
0
,
−
800
(
k
+
1
)
x
2
+
12
x
=
0
12
x
3
=
800
(
k
+
1
)
x
3
=
200
3
(
k
+
1
)
\begin{gather*} \frac{-800(k+1)}{x^2} + 12x = 0 \\ 12x^3 = 800(k+1) \\ x^3 = \frac{200}{3}(k+1) \\ \end{gather*}
x
2
−
800
(
k
+
1
)
+
12
x
=
0
12
x
3
=
800
(
k
+
1
)
x
3
=
3
200
(
k
+
1
)
x
=
200
3
(
k
+
1
)
3
■
x = \sqrt[3]{\frac{200}{3}(k+1)} \; \blacksquare
x
=
3
3
200
(
k
+
1
)
■
d
2
A
d
x
2
=
1600
(
k
+
1
)
x
3
+
12
\frac{\mathrm{d}^{2}A}{\mathrm{d}x^{2}} = \frac{1600(k+1)}{x^3} + 12
d
x
2
d
2
A
=
x
3
1600
(
k
+
1
)
+
12
When
x
=
200
3
(
k
+
1
)
3
,
{x=\sqrt[3]{\frac{200}{3}(k+1)}, }
x
=
3
3
200
(
k
+
1
)
,
d
2
A
d
x
2
>
0
{\frac{\mathrm{d}^{2}A}{\mathrm{d}x^{2}}>0}
d
x
2
d
2
A
>
0
so
A
{A}
A
is a minimum
■
{\blacksquare}
■
(ii)
Using
(
1
)
,
{(1),}
(
1
)
,
y
x
=
100
x
3
=
100
200
3
(
k
+
1
)
=
3
2
(
k
+
1
)
■
\begin{align*} \frac{y}{x} &= \frac{100}{x^3} \\ &= \frac{100}{\frac{200}{3}(k+1)} \\ &= \frac{3}{2(k+1)} \; \blacksquare \end{align*}
x
y
=
x
3
100
=
3
200
(
k
+
1
)
100
=
2
(
k
+
1
)
3
■
(iii)
0
<
k
≤
1
1
<
k
+
1
≤
2
2
<
2
(
k
+
1
)
≤
4
3
4
≤
3
2
(
k
+
1
)
<
3
2
\begin{gather*} 0 < k \leq 1 \\ 1 < k+1 \leq 2 \\ 2 < 2(k+1) \leq 4 \\ \frac{3}{4} \leq \frac{3}{2}(k+1) < \frac{3}{2} \end{gather*}
0
<
k
≤
1
1
<
k
+
1
≤
2
2
<
2
(
k
+
1
)
≤
4
4
3
≤
2
3
(
k
+
1
)
<
2
3
3
4
≤
y
x
<
3
2
■
\frac{3}{4} \leq \frac{y}{x} < \frac{3}{2} \; \blacksquare
4
3
≤
x
y
<
2
3
■
(iv)
If the box has square ends,
y
=
x
y
x
=
1
3
2
(
k
+
1
)
=
1
k
+
1
=
3
2
k
=
1
2
■
\begin{align*} y &= x \\ \frac{y}{x} &= 1 \\ \frac{3}{2(k+1)} &= 1 \\ k+1 &= \frac{3}{2} \\ k &= \frac{1}{2} \; \blacksquare \end{align*}
y
x
y
2
(
k
+
1
)
3
k
+
1
k
=
x
=
1
=
1
=
2
3
=
2
1
■
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