




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} + t^{1/3}
and the minimum occurs at t = 2:
y_{min} = 2^{2/3} + 2^{1/3}
t will give rise to a value of y within 10% of the minimum so long as we have
t^{2/3} + t^{1/3} £ 1.1_{ymin}^{ } = 1.1(2^{2/3}+2^{1/3})
and therefore for t between the two solutions of the equation
t^{2/3} + t^{1/3} = 1.1(2^{2/3} + 2^{1/3})
We can solve this equation by Newton's method to find
t_{1} = 0.8
t_{2} = 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} + x^{1/3} 1.1(2^{2/3} + 2^{1/3}).
We choose, for example, x_{1} = 1
This yields the first solution: t_{1} = 0.8
We pick another point, let say x_{2} = 10
This yields the second solution: t_{2} = 5.39