# S(t) = −0.00003237t^5 + 0.0009037t^4 − 0.008956t^3 + 0.03629t^2 − 0.04527t + 0.4395A model for the average price of a pound of white sugar in a certain country from August 1993 to August...

###### S(*t*) = −0.00003237*t^*5 + 0.0009037*t*^4 − *0.008956t^3* + 0.03629*t^2* − 0.04527*t* + 0.4395

A model for the average price of a pound of white sugar in a certain country from August 1993 to August 2003 is given by the function where *t* is measured in years since August of 1993. Estimate the times when sugar was cheapest and most expensive during the period 1993-2003.

### 1 Answer | Add Yours

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. **