You should notice that this is an optimization problem, hence you should use derivative of function to find the minimum and maximum values.

You need to differentiate the function with respect to t such that:

`S'(t) = (−0.00003237t^5 + 0.0009037t^4 − 0.008956t^3 + 0.03629t^2 − 0.04527t + 0.4395)'`

`S'(t) = −5*0.00003237t^4 + 4*0.0009037t^3 - 3*0.008956t^2 + 2*0.03629t - 0.04527`

`S'(t) = −0.00016185t^4 + 0.0036148t^3 - 0.026868 t^2 + 0.027258t - 0.04527`

`S'(t) = −16185t^4/100000000 + 36148t^3/10000000- 26868t^2/1000000 + 27258t/1000000 - 0.04527`

`S'(t) = −16185t^4 + 361480t^3 - 2686800t^2 + 2725800t - 4527000`

You should solve the equation `S'(t) = 0` such that:

`-16185t^4 + 361480t^3 - 2686800t^2 + 2725800t - 4527000 = 0`

`-16185(t^2- 21.46t+145.309)(t^2-0.875t + 1.9) = 0`

`t^2- 21.46t+145.309 = 0`

`t_(1,2) = (21.46 +- sqrt(460.5316 - 581.236))/2`

`t_(1,2) = (21.46 +-10.98i)/2`

`t_(1,2) = 10.73 +- 5.493i`

`t^2-0.875t + 1.9 = 0`

`t_(1,2) = (0.875 +- sqrt(0.765625 - 7.6))/2`

`t_(1,2) = (0.875 +-2.614i)/2`

`t_(1,2) = 0.4375 +- 1.307i`

**Hence, since the roots of derivative are complex, there are not critical values for function and the function has not minimum or maximum points, thus it is not possible to tell when sugar was cheapest and most expensive during the period 1993-2003. **

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