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Application to tin cans

In the tin can example, consider the question:

What values of t give rise to areas within 10% of the minimum?

We have seen that in this example, we minimize the function y:

y = t-2/3 + t1/3

and the minimum occurs at t = 2:

ymin = 2-2/3 + 21/3

t will give rise to a value of y within 10% of the minimum so long as we have

t-2/3 + t1/3 1.1ymin = 1.1(2-2/3+21/3)

and therefore for t between the two solutions of the equation

t-2/3 + t1/3 = 1.1(2-2/3 + 21/3)

We can solve this equation by Newton's method to find

t1 = 0.8

t2 = 5.4

and any value between these gives an area within 10% of the minimum.

 

Solutions

Use the Newton's Method solver with

f(x) = x-2/3 + x1/3 -1.1(2-2/3 + 21/3).

We choose, for example, x1 = 1

This yields the first solution: t1 = 0.8

We pick another point, let say x2 = 10

This yields the second solution: t2 = 5.39