You need to use the derivative of the function S(t) to decide when sugar was the most expensive or the cheapest, during the given period, such that:

`S'(t) = (-0.00003237 t^5 + 0.0009037 t^4- 0.008956 t^3 + 0.03629 t^2 - 0.04537 t + 0.5051)'`

`S'(t) = -0.00016185 t^4 + 0.0036148...

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You need to use the derivative of the function S(t) to decide when sugar was the most expensive or the cheapest, during the given period, such that:

`S'(t) = (-0.00003237 t^5 + 0.0009037 t^4- 0.008956 t^3 + 0.03629 t^2 - 0.04537 t + 0.5051)'`

`S'(t) = -0.00016185 t^4 + 0.0036148 t^3 - 0.026868 t^2 + 0.07258 t- 0.04537 `

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

`-0.00016185 t^4 + 0.0036148 t^3 - 0.026868 t^2 + 0.07258 t - 0.04537= 0`

Using the garph yields that the equation `S'(t) = 0` has 4 real roots `t_1=0.8, t_2=4.5, t_3=9.5, t_4=7.4` .

Notice that `S'(t)<0` for `tlt0.8` and `S'(t)>0` for `0.8<t<4.5` , hence, the sugar is the cheapest at `t=0.8` .

Notice that `S'(t)lt0` for `tgt4.5` , hence, the sugar is the most expensive at `t=4.5` .

Notice that `S'(t)<0` for`tgt7.4` and `S'(t)gt0` for `tgt7.4` , hence, the sugar is the cheapest at `t=7.4` .

Notice that `S'(t)<0` for `t>9.5` , hence, the sugar is the most expensive at `t=9.5` .

**Hence, the sugar is the most cheapest at `t=0.8` and `t=7.4` and it is the most expensive at `t=4.5` and `t=9.5` .**