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

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 + x^1/3 -1.1(2^-2/3 + 2^1/3).

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