Cylindrical Cans

00:00 15 marks

A company produces cylindrical cans for soup. The standard can has a radius \(r\) cm and a height \(h\) cm. The volume of the standard can is 450 cm³.

(a) Show that the external surface area, \(A\) cm², of the can is given by \(A = 2πr² + \frac{900}{r}\). [3]

(b) Find \(\frac{dA}{dr}\). [2]

The company wants to minimize the amount of metal used for the can, which means minimizing the external surface area.

(d) Calculate the minimum external surface area of the can. [2]

(e) Calculate the value of \(h\) corresponding to this minimum surface area. [2]

A different company produces novelty oversized cans where the height is always three times the radius.

(f) Find the radius of an oversized can that has the same volume (450 cm³) as the standard can. [3]

Hint Hint

Hint

Use the volume relation \(V = πr²h = 450\) to express \(h\) in terms of \(r\) and substitute into the surface area formula. Differentiate the resulting expression, set the derivative to zero, and solve for \(r\). Then substitute back to find \(h\) and similarly set up the volume equation for the oversized can.

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Complete Answer

a) Substitute \(h = 450/(πr²)\) into \(A = 2πr² + 2πrh\) to obtain \(A = 2πr² + 900/r\). b) Differentiate to get \(dA/dr = 4πr – 900/r²\). c) Setting \(4πr – 900/r² = 0\) leads to \(r³ = 225/π\), so \(r = \sqrt[3]{225/π} \approx 4.15\) cm. d) Substituting \(r \approx 4.15\) into the area formula gives \(A \approx 325\) cm². e) Using \(h = 450/(πr²)\) with \(r \approx 4.15\) results in \(h \approx 8.30\) cm. f) For an oversized can where \(h = 3r\), the volume equation \(3πr³ = 450\) yields \(r³ = 150/π\), so \(r = \sqrt[3]{150/π} \approx 3.63\) cm.

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