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H1 Math

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PSLE

PSLE Foundation Mathematics

<p>Rectangle $OPQR$ has sides parallel to the coordinate axes, with $O$ at the origin, $P$ on the positive $x$-axis and $R$ on the positive $y$-axis. The vertex $Q(x,y)$ lies in the first quadrant on the curve $C$, whose equation is $x^2+2xy+4y^2=52$. The variables $x$ and $y$ are measured in units. Let&nbsp;$A$ be the area of the rectangle.</p>
<div class="graph-block" data-graph-config="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" data-width="300" data-height="200" style="width:300px;height:200px;max-width:100%;display:block;margin:0 auto;"></div>
<ol class="i-list"><li data-list-item-id="implicit-rectangle-part-i" data-list-item-path="implicit-rectangle-part-i"><div class="list-item-paragraph">Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$ and $y$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="implicit-rectangle-part-ii" data-list-item-path="implicit-rectangle-part-ii"><div class="list-item-paragraph">Hence, show that&nbsp;$A$ has a stationary value when&nbsp;$x=2y$, and find the exact maximum value of $A$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="implicit-rectangle-part-iii" data-list-item-path="implicit-rectangle-part-iii"><div class="list-item-paragraph">At an instant, $Q=(6,1)$ and $x$ is increasing at a constant rate of $\frac{5}{2}$ units/s. Find the rate of change of $A$ at this instant.<span class="inline-marks">[3]</span></div></li></ol>

Rectangle $OPQR$ has sides parallel to the coordinate axes, with $O$ at the origin, $P$ on the positive $x$-axis and $R$ on the positive $y$-axis. The vertex $Q(x,y)$ lies in the first quadrant on the curve $C$, whose equation is $x^2+2xy+4y^2=52$. The variables $x$ and $y$ are measured in units. Let $A$ be the area of the rectangle.

Implicit Differentiation

Connected Rates of Change

Problems Involving Maximisation and Minimisation

Rectangle under an implicit curve

Applications of Differentiation

Techniques of Differentiation

<p>An isosceles triangle $ABC$, with $AB = AC$, is inscribed in a fixed circle of radius 1 unit and centre $O$. It is given that angle $BOC = 2\theta$, where&nbsp;$\theta$ is acute. Using calculus, show that the area of the triangle $ABC$ is a maximum when it is equilateral. State the maximum value of this area exactly.<span class="inline-marks">[8]</span></p>

An isosceles triangle $ABC$, with $AB = AC$, is inscribed in a fixed circle of radius 1 unit and centre $O$. It is given that angle $BOC = 2\theta$, where $\theta$ is acute. Using calculus, show that the area of the triangle $ABC$ is a maximum when it is equilateral. State the maximum value of this area exactly.

2024 NYJC P1 Q7

<p>[The volume of a square-based pyramid is $\frac{1}{3} \times \text{base area} \times \text{height}$ .]</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1779693069635-n5rr3e4b31.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Nzk2OTMwNjk2MzUtbjVycjNlNGIzMS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTU2MzUwLCJleHAiOjE3OTM1NTY2NTB9.bWd5OmSdqcnf_k4xNK6W7jf4gKPFOvyDViJr-RjciNs" style="max-width: min(150px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>A manufacturer makes open containers in the shape of an inverted square pyramid with square base of side length $l$ cm, vertical height $h$ cm and fixed slant edge $k$ cm, as shown above. He decides that the volume of each container, $V\text{ cm}^3$, should be as large as possible.</p>
<ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-paragraph">Show that&nbsp;$V^2 = \frac{k^2l^4}{9} - \frac{l^6}{18}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b"><div class="list-item-paragraph">Use differentiation to find, in terms of $k$, the exact maximum value of $V$, proving that it is a maximum.<span class="inline-marks">[6]</span></div></li><li data-list-item-id="item-c"><div class="list-item-paragraph">Water is poured into an empty container with the dimensions found in part <strong>(b)</strong> at a constant rate of $\frac{1}{3}\text{ cm}^3$ per second. It is given that at time $t$ seconds, the volume of water in the container is $W\text{ cm}^3$ and the water level is $p$ cm. Show that&nbsp;$W = \frac{4}{3}p^3$ and hence, find the rate of increase of $p$ at the instant when&nbsp;$p = \frac{k}{4}$.<span class="inline-marks">[4]</span></div></li></ol>

[The volume of a square-based pyramid is $\frac{1}{3} \times \text{base area} \times \text{height}$ .]

A manufacturer makes open containers in the shape of an inverted square pyramid with square base of side length $l$ cm, vertical height $h$ cm and fixed slant edge $k$ cm, as shown above. He decides that the volume of each container, $V\text{ cm}^3$, should be as large as possible.

2024 VJC Promo Q12

<p>A curve $C$ has parametric equations</p>
<p style="text-align: center">\[x = \frac{\lambda}{1 + \lambda^3} \quad \text{and} \quad y = \frac{\lambda^2}{1 + \lambda^3}, \quad \text{for}\hspace{0.5em} \lambda &gt; 0.\]</p>
<p>The point $P$ is a variable point on $C$.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">With reference to the origin $O$, $OP$ forms the diagonal of the rectangle $OQPR$, where vertices $Q$ and $R$ lie on the $x$- and $y$-axis respectively. Using differentiation, find the value of $\lambda$ which maximises the area of rectangle $OQPR$. You need to show that your answer gives a maximum.<span class="inline-marks">[5]</span></div></li></ol>
<p>When the area of rectangle $OQPR$ is at its maximum, the rate of change of the $x$-coordinate of the point $P$ is $1$ unit per second.</p>
<ol class="abc-list" start="2"><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ and hence determine the rate of change of the $y$-coordinate of the point $P$ at this instant.<span class="inline-marks">[4]</span></div></li></ol>

\[x = \frac{\lambda}{1 + \lambda^3} \quad \text{and} \quad y = \frac{\lambda^2}{1 + \lambda^3}, \quad \text{for } \lambda > 0.\]

The point $P$ is a variable point on $C$.

When the area of rectangle $OQPR$ is at its maximum, the rate of change of the $x$-coordinate of the point $P$ is $1$ unit per second.

Quotient Rule

Differentiation of Parametric Equations

2024 RI Promo Q4

<p>A closed triangular prism has length $l$ cm and an equilateral triangle of side $x$ cm as its base.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1779549311101-m51g4gllje.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Nzk1NDkzMTExMDEtbTUxZzRnbGxqZS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTU2MzUwLCJleHAiOjE3OTM1NTY2NTB9.FncT6u11ZtuRQCQ_AlgNS67leHdzWxHb8ned4xTp2Vo" style="max-width: min(250px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-paragraph">Given that the total surface area of this prism is 150 cm${}^2$, show that the volume $V$ cm${}^3$ of the prism is given by</div><div class="list-item-paragraph" style="text-align: center">\[V = \frac{1}{8}x\left(100\sqrt{3} - x^2\right).\]<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-ii"><div class="list-item-paragraph">Hence using differentiation, find the value of the maximum volume of the prism, proving that it is a maximum.<span class="inline-marks">[4]</span></div></li></ol>

A closed triangular prism has length $l$ cm and an equilateral triangle of side $x$ cm as its base.

2024 MI PRU2 P1 Q6

<p>The diagram shows an ellipse, centred at the origin with semi-minor axis $a$ and semi-major axis $2a$, where&nbsp;$a &gt; 0$. The rectangle $PQRS$ is inscribed in the ellipse such that its four vertices $P$, $Q$, $R$ and $S$ lie on the ellipse. The coordinates of $P$ is $(x, y)$, where&nbsp;$x$ and $y$ are positive.</p>
<div class="graph-block" data-graph-config="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" data-width="250" data-height="250" style="width:250px;height:250px;max-width:100%;display:block;margin:0 auto;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Write down the cartesian equation of the ellipse in terms of $a$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">By considering the coordinates of $P$, show that&nbsp;$A$, the area of rectangle $PQRS$, is $8x\sqrt{a^2 - x^2}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Using differentiation, find, in terms of $a$, the value of $x$ when&nbsp;$A$ is a maximum. You do not need to prove that&nbsp;$A$ is a maximum.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">Find the value of $a$ if the maximum value of $A$ is $100$.<span class="inline-marks">[2]</span></div></li></ol>

The diagram shows an ellipse, centred at the origin with semi-minor axis $a$ and semi-major axis $2a$, where $a > 0$. The rectangle $PQRS$ is inscribed in the ellipse such that its four vertices $P$, $Q$, $R$ and $S$ lie on the ellipse. The coordinates of $P$ is $(x, y)$, where $x$ and $y$ are positive.

Graphs of Conic Sections

2024 JPJC Promo Q6

Graphing Techniques

<p>Mrs Koh wants to hang a decoration on the vertical wall above her bookshelf. She needs a ladder to climb up.</p>
<p>The rectangle $ABCD$ is the side-view of the bookshelf and $HK$ is the side-view of the ladder where&nbsp;$AB = 24\text{ cm}$ and $BC = 192\text{ cm}$ (see <strong>Figure 1</strong>). The ladder touches the wall at $H$, the edge of the top of the bookshelf at $B$ and the floor at $K$.</p>
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<ol class="abc-list"><li><div class="list-item-paragraph">Given that&nbsp;$\angle HKD = \theta$, show that the length, $L\text{ cm}$ of the ladder is given by \[L = \frac{24}{\cos \theta} + \frac{192}{\sin \theta}.\]<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Use differentiation to find the exact value of the shortest length of the ladder as&nbsp;$\theta$ varies.<br>[You do not need to verify that this length of the ladder is the shortest.]<span class="inline-marks">[5]</span></div></li></ol>

Mrs Koh wants to hang a decoration on the vertical wall above her bookshelf. She needs a ladder to climb up.

The rectangle $ABCD$ is the side-view of the bookshelf and $HK$ is the side-view of the ladder where $AB = 24\text{ cm}$ and $BC = 192\text{ cm}$ (see **Figure 1**). The ladder touches the wall at $H$, the edge of the top of the bookshelf at $B$ and the floor at $K$.

Derivatives of Reciprocal Trigonometric Functions

<p>A curve has equation $y = \frac{x^{n+1}}{n+1} + \frac{a}{x}$ for some non-zero constants $a$ and $n$, with $n\ne-1$. The curve has a common gradient value of $\frac{5}{2}$ at both&nbsp;$x = \sqrt{2}$ and $x = \frac{1}{\sqrt{2}}$.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">By using the substitution $b = 2^{\frac{n}{2}}$, find the equation of the curve.<span class="inline-marks">[7]</span></div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">By considering the expansion of $\left( x - \frac{1}{x} \right)^2$, show that the gradient of the curve at all points on the curve is at least $2$.<span class="inline-marks">[3]</span></div></li></ol>

A curve has equation $y = \frac{x^{n+1}}{n+1} + \frac{a}{x}$ for some non-zero constants $a$ and $n$, with $n\ne-1$. The curve has a common gradient value of $\frac{5}{2}$ at both $x = \sqrt{2}$ and $x = \frac{1}{\sqrt{2}}$.

By using the substitution $b = 2^{\frac{n}{2}}$, find the equation of the curve.

By considering the expansion of $\left( x - \frac{1}{x} \right)^2$, show that the gradient of the curve at all points on the curve is at least $2$.

Derivative as the Gradient of a Curve

NJC IP4 Techniques of Differentiation Tutorial Q1

Differentiation

<p>The diagram shows a trapezium $OPQR$, where&nbsp;$OR = 6$ cm and angle $OPQ = \text{angle}\hspace{0.5em} PQR = \frac{\pi}{2}$ radians. Angle $ROP = \theta$ radians, where&nbsp;$\theta$ is acute. $T$ is a point on $PQ$ such that angle $ORT = \frac{\pi}{2}$ radians and $RT = 3$ cm.</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Show that the area of the trapezium $OPQR$, in cm$^2$, is $9 \sin 2\theta + 18 \sin^2 \theta$.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Given that&nbsp;$\theta$ can vary, find the exact value of $\theta$ for which the area of trapezium $OPQR$ is stationary.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Determine whether this value of $\theta$ gives a maximum or minimum value of the area.</div></li></ol>

The diagram shows a trapezium $OPQR$, where $OR = 6$ cm and angle $OPQ = \text{angle } PQR = \frac{\pi}{2}$ radians. Angle $ROP = \theta$ radians, where $\theta$ is acute. $T$ is a point on $PQ$ such that angle $ORT = \frac{\pi}{2}$ radians and $RT = 3$ cm.

Show that the area of the trapezium $OPQR$, in cm$^2$, is $9 \sin 2\theta + 18 \sin^2 \theta$.

Given that $\theta$ can vary, find the exact value of $\theta$ for which the area of trapezium $OPQR$ is stationary.

Determine whether this value of $\theta$ gives a maximum or minimum value of the area.

Derivatives of Trigonometric Functions

<p>Figure I shows a shelter made by a folded rectangular metal sheet of a fixed area of $1080$ m$^2$, where the length of the shelter is $60$ m. Figure II shows the cross section of the shelter, where&nbsp;$ABDE$ is a rectangle with $AB = DE = x$ m and triangle $BCD$ is isosceles with $BC = CD = 2x$ m. Let angle $CBD =$ angle $CDB = \theta$ radians.</p>
<div class="graph-block" 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data-width="542" data-height="219" data-proportional style="width:542px;height:219px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:542px;--graph-aspect:542 / 219;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">State the range of values of $\theta$.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Find the value of $x$.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Show that the internal volume of the shelter, $V$ m$^3$, is given by $V = 1080(\sin 2\theta + 2\cos \theta)$.</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">Find the exact value of $\theta$ for which $V$ has a stationary value.</div></li><li data-list-item-id="item-v" data-list-item-path="item-v"><div class="list-item-paragraph">Hence find the stationary value of $V$ and determine its nature. Give the exact value.</div></li></ol>

Figure I shows a shelter made by a folded rectangular metal sheet of a fixed area of $1080$ m$^2$, where the length of the shelter is $60$ m. Figure II shows the cross section of the shelter, where $ABDE$ is a rectangle with $AB = DE = x$ m and triangle $BCD$ is isosceles with $BC = CD = 2x$ m. Let angle $CBD =$ angle $CDB = \theta$ radians.

Show that the internal volume of the shelter, $V$ m$^3$, is given by $V = 1080(\sin 2\theta + 2\cos \theta)$.

Find the exact value of $\theta$ for which $V$ has a stationary value.

Hence find the stationary value of $V$ and determine its nature. Give the exact value.

<p>The diagram shows an inverted cone with slant height $3$ cm and semi-vertical angle $\theta$ radians.</p>
<div class="graph-block" data-graph-config="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" data-width="137" data-height="176" data-proportional style="width:137px;height:176px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:137px;--graph-aspect:137 / 176;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">State the range of values of $\theta$.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Show that the volume of the cone, $V$ cm$^3$, is given by $V = 9\pi \sin^2 \theta \cos \theta$.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Find the values of $\sin \theta$ and $\cos \theta$ for which $V$ has a stationary value.</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">Hence find the stationary value of $V$ and determine its nature. Leave your answer in terms of $\pi$.</div></li></ol>

The diagram shows an inverted cone with slant height $3$ cm and semi-vertical angle $\theta$ radians.

Show that the volume of the cone, $V$ cm$^3$, is given by $V = 9\pi \sin^2 \theta \cos \theta$.

Find the values of $\sin \theta$ and $\cos \theta$ for which $V$ has a stationary value.

Hence find the stationary value of $V$ and determine its nature. Leave your answer in terms of $\pi$.

<p>The diagram shows a semicircle construction $ABCD$ with radius $r$ cm and centre $O$. The supporting structure consists of trapezium $ABCD$, radii $OB$ and $OC$. Trapezium $ABCD$ is such that&nbsp;$BC$ is parallel to $AD$ and has a height of $h$ cm. Let angle $ABO = \theta$ radians and $BC = 2x$ cm.</p>
<div class="graph-block" 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data-width="399" data-height="216" data-proportional style="width:399px;height:216px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:399px;--graph-aspect:399 / 216;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">State the range of values of $\theta$.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Express $x$ and $h$ in terms of $\theta$ and $r$.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Hence show that the shaded area, $A$ cm$^2$, is given by $A = \frac{1}{2} r^2 (\pi - 2 \sin 2\theta + \sin 4\theta).$</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">Given that&nbsp;$\theta$ can vary, find the value of $\theta$ for which $A$ has a stationary value.</div></li><li data-list-item-id="item-v" data-list-item-path="item-v"><div class="list-item-paragraph">Hence find the stationary value of $A$ in terms of $r$ and $\pi$, and determine its nature.</div></li></ol>

The diagram shows a semicircle construction $ABCD$ with radius $r$ cm and centre $O$. The supporting structure consists of trapezium $ABCD$, radii $OB$ and $OC$. Trapezium $ABCD$ is such that $BC$ is parallel to $AD$ and has a height of $h$ cm. Let angle $ABO = \theta$ radians and $BC = 2x$ cm.

Express $x$ and $h$ in terms of $\theta$ and $r$.

Given that $\theta$ can vary, find the value of $\theta$ for which $A$ has a stationary value.

Hence find the stationary value of $A$ in terms of $r$ and $\pi$, and determine its nature.

<p>The diagram shows the vertical cross-section $PQRS$ of an open trough. The lengths of $PQ$, $QR$ and $RS$ are $30$ cm, $40$ cm and $30$ cm, respectively. The trough rests with $QR$ as the base on horizontal ground and both&nbsp;$PQ$ and $RS$ are inclined at an acute angle $\theta$ radians to the ground.</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Show that the area, $A$ $\text{cm}^2$, of the cross-section $PQRS$ is given by $A = 1200 \sin \theta + 450 \sin 2\theta$.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Given that&nbsp;$\theta$ can vary, find the value of $\theta$ for which the volume of the trough is maximised.</div></li></ol>

The diagram shows the vertical cross-section $PQRS$ of an open trough. The lengths of $PQ$, $QR$ and $RS$ are $30$ cm, $40$ cm and $30$ cm, respectively. The trough rests with $QR$ as the base on horizontal ground and both $PQ$ and $RS$ are inclined at an acute angle $\theta$ radians to the ground.

Show that the area, $A$ $\text{cm}^2$, of the cross-section $PQRS$ is given by $A = 1200 \sin \theta + 450 \sin 2\theta$.

Given that $\theta$ can vary, find the value of $\theta$ for which the volume of the trough is maximised.

<p>A square-based prism fish tank with no lid has a fixed volume of $2000\text{ cm}^3$. The cost of the materials to manufacture this fish tank is such that, the cost of the base per $\text{cm}^2$ is twice the cost of the side faces per $\text{cm}^2$. Assume that the side faces cost $\$ 1$ per $\text{cm}^2$ and that the thickness of the tank is negligible. Let&nbsp;$x\text{ cm}$ be the length of the side of the base square.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find the value of $x$ for which the total cost of manufacturing the fish tank is a minimum.</div></li></ol>
<p>Another fish tank of different shape is initially empty. Water is pumped in the fish tank at a constant rate of $4\text{ cm}^3/\text{s}$. Let&nbsp;$h\text{ cm}$ be the depth and $V\text{ cm}^3$ be the volume of water in the tank at any time. It is given that&nbsp;$V = (2h^2 + h)^{\frac{2}{3}}$.</p>
<ol class="i-list" start="2"><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Find the rate of change of $h$ when&nbsp;$h = 10\text{ cm}$.</div></li></ol>

A square-based prism fish tank with no lid has a fixed volume of $2000\text{ cm}^3$. The cost of the materials to manufacture this fish tank is such that, the cost of the base per $\text{cm}^2$ is twice the cost of the side faces per $\text{cm}^2$. Assume that the side faces cost $\$ 1$ per $\text{cm}^2$ and that the thickness of the tank is negligible. Let $x\text{ cm}$ be the length of the side of the base square.

Find the value of $x$ for which the total cost of manufacturing the fish tank is a minimum.

Another fish tank of different shape is initially empty. Water is pumped in the fish tank at a constant rate of $4\text{ cm}^3/\text{s}$. Let $h\text{ cm}$ be the depth and $V\text{ cm}^3$ be the volume of water in the tank at any time. It is given that $V = (2h^2 + h)^{\frac{2}{3}}$.

Find the rate of change of $h$ when $h = 10\text{ cm}$.

<p>The diagram shows an inverted right cone with a base radius of $6$ cm and a height of $18$ cm, where a cylinder with a base radius of $r$ cm is placed upright inside the cone, such that the top of the cylinder is $4$ cm above the top of the cone. Let the part of the cylinder inside the cone be $h$ cm in depth.</p>
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data-width="227" data-height="244" data-proportional style="width:227px;height:244px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:227px;--graph-aspect:227 / 244;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Express $h$ in terms of $r$.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Find, in terms of $r$, the volume of the cylinder.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Find the value of $r$ for which the volume of the cylinder has a stationary value and determine the nature of this stationary value.</div></li></ol>

The diagram shows an inverted right cone with a base radius of $6$ cm and a height of $18$ cm, where a cylinder with a base radius of $r$ cm is placed upright inside the cone, such that the top of the cylinder is $4$ cm above the top of the cone. Let the part of the cylinder inside the cone be $h$ cm in depth.

Find, in terms of $r$, the volume of the cylinder.

Find the value of $r$ for which the volume of the cylinder has a stationary value and determine the nature of this stationary value.

<p>The diagram shows a sphere with a fixed radius $a$ cm and centre $O$ which is inscribed exactly inside a right cone of base radius $r$ cm and height $h$ cm</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Express $r$ in terms of $a$ and $h$.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">The volume of cone is $V$ cm$^3$. Show that&nbsp;$V = \frac{\pi a^2 h^2}{3(h - 2a)}$.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Find the value of $h$ for which $V$ has a turning value and determine the nature of this turning value.</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">Hence find this volume, leaving your answer in terms of $\pi$.</div></li></ol>

The diagram shows a sphere with a fixed radius $a$ cm and centre $O$ which is inscribed exactly inside a right cone of base radius $r$ cm and height $h$ cm

Express $r$ in terms of $a$ and $h$.

The volume of cone is $V$ cm$^3$. Show that $V = \frac{\pi a^2 h^2}{3(h - 2a)}$.

Find the value of $h$ for which $V$ has a turning value and determine the nature of this turning value.

Hence find this volume, leaving your answer in terms of $\pi$.

<p>The diagram shows a right cone with a base radius $r$ cm and height $h$ cm which is inscribed exactly inside a sphere of radius $6$ cm and centre $O$.</p>
<div class="graph-block" data-graph-config="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" data-width="237" data-height="204" data-proportional style="width:237px;height:204px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:237px;--graph-aspect:237 / 204;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Express $r$ in terms of $h$.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">The volume of cone is $V$ cm$^3$. Show that&nbsp;$V = 4\pi h^2 - \frac{1}{3}\pi h^3$.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Find the value of $r$ for which $V$ has a turning value and determine the nature of this turning value.</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">Hence find this volume, leaving your answer in terms of $\pi$.</div></li></ol>

The diagram shows a right cone with a base radius $r$ cm and height $h$ cm which is inscribed exactly inside a sphere of radius $6$ cm and centre $O$.

The volume of cone is $V$ cm$^3$. Show that $V = 4\pi h^2 - \frac{1}{3}\pi h^3$.

Find the value of $r$ for which $V$ has a turning value and determine the nature of this turning value.

<p>A factory manufactures cylindrical containers with fixed volume of $\frac{5\pi}{128}$ cm$^3$. The material used for the top and the base of the cylinder costs $\$ \ a$ / cm$^2$ and the material used for the curved surface costs $\$ \ \frac{4a}{5}$ / cm$^2$. Assume that the thickness of the container is negligible.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Given that the container has a base radius of $r$ cm, which can vary, find the total cost of manufacturing one such container in terms of $r$.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Hence find the value of $r$ for which the total cost has a minimum value.</div></li></ol>

A factory manufactures cylindrical containers with fixed volume of $\frac{5\pi}{128}$ cm$^3$. The material used for the top and the base of the cylinder costs $\$ \ a$ / cm$^2$ and the material used for the curved surface costs $\$ \ \frac{4a}{5}$ / cm$^2$. Assume that the thickness of the container is negligible.

Given that the container has a base radius of $r$ cm, which can vary, find the total cost of manufacturing one such container in terms of $r$.

Hence find the value of $r$ for which the total cost has a minimum value.

<p>A string of fixed length $l$ m is cut into $3$ segments to enclose a square, equilateral triangle and a circle, where the length of sides of the square and the equilateral triangle is $x$ m and the radius of the circle is $r$ m.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Express $r$ in terms of $x$ and $l$.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Show that the total area of the $3$ enclosed figures, $A$ $\text{m}^2$, is given by<br>$A = \left( \frac{\sqrt{3}}{4} + 1 \right)x^2 + \frac{(l - 7x)^2}{4\pi} .$</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Given that&nbsp;$x$ can vary, find the value of $x$ for which $A$ has a minimum value.</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">Hence find the ratio of the lengths of the $3$ segments cut out of the original string.</div></li></ol>

A string of fixed length $l$ m is cut into $3$ segments to enclose a square, equilateral triangle and a circle, where the length of sides of the square and the equilateral triangle is $x$ m and the radius of the circle is $r$ m.

Express $r$ in terms of $x$ and $l$.

Show that the total area of the $3$ enclosed figures, $A$ $\text{m}^2$, is given by\n\n\[A = \left( \frac{\sqrt{3}}{4} + 1 \right)x^2 + \frac{(l - 7x)^2}{4\pi} .\]

Given that $x$ can vary, find the value of $x$ for which $A$ has a minimum value.

Hence find the ratio of the lengths of the $3$ segments cut out of the original string.

<p>As shown in the diagram, a person in a boat is $3$ km from the nearest point from the straight beach. The person is to get to a cabin $6$ km along the beach. It is known that the boat travels at $4$ km/h and the person walks at $5$ km/h. Assume that the person lands at a point $x$ km from the point directly opposite to the original position.</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find, in terms of $x$, the total time (in hours) taken for the person to reach the cabin.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Find the value of $x$ for which the total time taken is the minimum and find this minimum time.</div></li></ol>

As shown in the diagram, a person in a boat is $3$ km from the nearest point from the straight beach. The person is to get to a cabin $6$ km along the beach. It is known that the boat travels at $4$ km/h and the person walks at $5$ km/h. Assume that the person lands at a point $x$ km from the point directly opposite to the original position.

Find, in terms of $x$, the total time (in hours) taken for the person to reach the cabin.

Find the value of $x$ for which the total time taken is the minimum and find this minimum time.

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