lab10-extra

New topic: optimisation (optimal_can)

optimal_can(volume=330)

Minimise can surface area for fixed volume.

A cylindrical drinks can has radius \(r\) and height \(h\). Its volume is

\[V = V(r, h) = \pi r^2 h\]

and its total surface area (top + bottom + side wall) is

\[A = A(r, h) = 2\pi rh + 2\pi r^2.\]

Write two functions:

• cylinder_area_from_radius_and_volume(radius, volume)

• optimal_can(volume)

For optimal_can(volume), return (diameter, height) for the minimum-area can with that fixed volume volume.

If no volume is provided, assume the default value of 330.

Use scipy.optimize.fmin to solve the optimisation.

Hints:

• In cylinder_area_from_radius_and_volume, use the volume relation to eliminate h:

  \[h = \frac{V}{\pi r^2}\]

  and substitute into

  \[A = 2\pi rh + 2\pi r^2\]

  so the area A depends only on r for a given and fixed V.

• In optimal_can, minimise this one-variable area function cylinder_area_from_radius_and_volume with fmin. In other words, using fmin, find the value r for the radius, so that the area A is minimal.

• fmin returns a numpy array (of length 1); extract the optimal radius from it.

• The fmin function takes on optional keyword argument with name args. This is a list of tuples which will be passed onto the objective function. You need to use it, to pass the volume to cylinder_area_from_radius_and_volume. For example:

  args = (volume, )  # needs to be a tuple, even if it contains only one argument
  fmin(cylinder_area_from_radius_and_volume, r_initial, args=args)
  

• To convert the optimum radius \(r\) into height and diametre: \(h = V / (\pi r^2)\) and \(d = 2r\).

Examples

In [ ] : cylinder_area_from_radius_and_volume(1, 10)
Out[27]: 26.283185307179586

In [ ] : cylinder_area_from_radius_and_volume(2, 10)
Out[28]: 35.132741228718345

For this example we express the distances – such as the radius, diameter (\(d = 2r\)) and height – in centimetres (cm), and area in square cm and volumina in cubic cm. For 330 cm^3 and a call of optimal_can(330), both returned values for height and diametre should be close to about 7.5 (which then means 7.5 cm).

Voluntary follow up question (nothing to submit here): the standard 330cl drink can (for example from this provider) is slimmer: diametre is smaller than 7cm and height is above 11cm. Can you think why that may be the case? In addition to any PR-stunts, there is an engineering/economy reason.

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using_quad()

Numerical integration

Write a function using_quad() that computes and returns the integral \(\int\limits_a^b x^2 f(x) \mathrm{d}x\) with \(f(x) = x^2\) from \(a=0\) to \(b=2\) using scipy.integrate.quad. The function using_quad should return the value that the quad() function returns, i.e. a tuple (y, abserr) where y is quad’s best approximation of the true integral, and abserr an absolute error estimate.

You should find that this method (with the default settings) is (far) more accurate than our trapez function.

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Please include the extra tasks in your file lab10.py and submit as lab10 assignment.

Back to lab10.

End of lab10-extra.