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A soft drink can has a volume of 340 cm³, a height of h cm and a radius of r cm - NSC Mathematics - Question 9 - 2016 - Paper 1

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A soft drink can has a volume of 340 cm³, a height of h cm and a radius of r cm. 9.1 Express h in terms of r. 9.2 Show that the surface area of the can is given by... show full transcript

Worked Solution & Example Answer:A soft drink can has a volume of 340 cm³, a height of h cm and a radius of r cm - NSC Mathematics - Question 9 - 2016 - Paper 1

Step 1

Express h in terms of r.

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Answer

To express the height h in terms of the radius r, we can use the formula for the volume of a cylinder, given by:

V = ext{Base Area} imes ext{Height} = rac{22}{7}r^2h

Setting this equal to the given volume of 340 cm³, we have:

340 = rac{22}{7} r^2 h

Rearranging for h gives:

h = rac{340 imes 7}{22 r^2} = rac{2380}{22 r^2} = rac{109.09}{r^2}.

Step 2

Show that the surface area of the can is given by A(r) = 2πr² + 680r⁻¹.

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Answer

The surface area A of a cylinder can be given by:

A = 2 ext{Base Area} + ext{Lateral Surface Area} = 2 rac{22}{7}r^2 + 2 rac{22}{7}rh

Substituting h from the previous part, we have:

A = 2 rac{22}{7}r^2 + 2 rac{22}{7}r\left(\frac{2380}{22 r^2}\right)

This simplifies to:

A = 2 rac{22}{7}r^2 + rac{47600}{22 r}

Thus,

A(r) = 2 rac{22}{7}r^2 + 680r^{-1}.

Step 3

Determine the radius of the can that will ensure the surface area is a minimum.

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Answer

To find the radius that minimizes the surface area, we differentiate A with respect to r:

A'(r) = 4 rac{22}{7}r - 680r^{-2}

Setting the derivative equal to zero:

4 rac{22}{7}r - 680r^{-2} = 0

This leads to:

4 rac{22}{7}r^3 = 680

Solving for r:

r3=680imes74imes22r^3 = \frac{680 imes 7}{4 imes 22}

Calculating this gives:

r3=17.5r=17.532.6cmr^3 = 17.5 \Rightarrow r = \sqrt[3]{17.5} ≈ 2.6\,\text{cm}.

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