Estimate the times when sugar was cheapest and most expensive during the period 1993-2003. given by the function S(t) = −0.00003237t^5 + 0.0009037t^4 − 0.008956t^3+ 0.03629t^2− 0.04537t + 0.5051 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 S(t) = −0.00003237t^5 + 0.0009037t^4 − 0.008956t^3+ 0.03629t^2− 0.04537t + 0.5051 where t is measured in years since August of 1993. (Round your answers to three decimal places.) t=_________________(cheapest) t=_________________(most expensive)

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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...

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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` .

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