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.
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