A cylinder is to be cut out of the circular face of a solid hemisphere. The cylinder and the hemisphere have the same axis of symmetry. The cylinder has height $h$ and the hemisphere has a radius of $R$. 9 (a) Show that the volume, $V$, of the cylinder is given by $V = ho imes R^{2} imes h - rac{ heta}{3}$. 9 (b) Find the maximum volume of the cylinder in terms of $R$. Fully justify your answer.

A-Level AQA Maths Pure Further Applications of Differentiation: A cylinder is to be cut out of t

A cylinder is to be cut out of the circular face of a solid hemisphere - AQA - A-Level Maths Pure - Question 9 - 2020 - Paper 2

Question 9

A cylinder is to be cut out of the circular face of a solid hemisphere. The cylinder and the hemisphere have the same axis of symmetry. The cylinder has height $h$ a... show full transcript

Worked Solution & Example Answer:A cylinder is to be cut out of the circular face of a solid hemisphere - AQA - A-Level Maths Pure - Question 9 - 2020 - Paper 2

Step 1

Show that the volume, V, of the cylinder is given by

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Answer

To find the volume of the cylinder, we first identify the relevant variables. The volume $V$ of a cylinder can be expressed as:

$V = ext{Base Area} imes ext{Height}$

The base area of the cylinder is a circle with radius $r$ . Therefore:

$V = ext{Area of Base} imes h = ext{Area of } imes h = ext{Base Area} = au imes r^{2} imes h = au imes R^{2} - rac{ heta}{3}.$

Next, we need to express $r$ in terms of $h$ . Using the geometry of the situation:

From the right triangle formed, we apply the Pythagorean theorem:

$R^{2} = r^{2} + h^{2}$

Thus, we can derive $r^2$ as:

$r^2 = R^2 - h^2$

Substituting $r^2$ into the volume formula yields:

$V = au imes (R^{2} - h^{2}) imes h = au R^{2}h - rac{ heta}{3},$

which completes the proof.

Step 2

Find the maximum volume of the cylinder in terms of R.

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Answer

To find the maximum volume of the cylinder, we first take the derived volume formula:

$V = au R^{2}h - rac{ heta}{3}$

Next, we differentiate the volume with respect to height $h$ :

$\frac{dV}{dh} = \tau R^{2} - 3 \theta h^{2}$

Setting this equal to zero to find critical points for maximum volume:

$0 = \tau R^{2} - 3 \theta h^{2}$

This implies:

$h^2 = \frac{\tau R^{2}}{3\theta}$

Next, we also need to consider the values of $h$ . The maximum occurs either when $h = 0$ or $h = R$ , but we focus on the stationary points:

Substituting this $h$ back into the volume expression:

We arrive at the form:

$V = K R^{2} h - rac{ heta}{3}$

Hence, substituting the term in the formula:

$h = \frac{R}{\sqrt{3}}$

Finally, to verify the maximum, substitute $h$ back into the differentiative volume formula:

$\frac{d^2V}{dh^2} = -6\theta < 0,$

showing that this indeed provides a maximum volume.

A cylinder is to be cut out of the circular face of a solid hemisphere - AQA - A-Level Maths Pure - Question 9 - 2020 - Paper 2

Worked Solution & Example Answer:A cylinder is to be cut out of the circular face of a solid hemisphere - AQA - A-Level Maths Pure - Question 9 - 2020 - Paper 2

Show that the volume, V, of the cylinder is given by

Find the maximum volume of the cylinder in terms of R.

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