Consider the given function\(s(t) = - 0.00003237{t^5} + 0.0009037{t^4} - 0.008956{t^3} + 0.03629{t^2} - 0.04458 + 0.4074\).
Differentiate the given function with respect to\(t\).
\({s^\prime }(t) = - 0.00016185{t^4} + 0.0036148{t^3} - 0.026868{t^2} + 0.07258t - 0.04458\)
Now, equate\({s^\prime }(t) = 0\)and evaluate the value of\(t\)by the use of a calculator.
\(\begin{aligned}{c}0 &= - 0.00016185{t^4} + 0.0036148{t^3} - 0.026868{t^2} + 0.07258t - 0.04458\\t &= 0.85478,4.6178,7.2919,9.5699\end{aligned}\)
The cheapest and the most expensive cost is evaluated by the substitution of the values of\(t\)obtained above in\(s(t) = 0\).
\(\begin{aligned}{c}s(0.85478) = 0.39068\\s(4.6178) = 0.43645\\s(7.2919) = 0.42712\\s(9.5699) = 0.43641\end{aligned}\)
From the above calculation it can be inferred that when\(t = 0.85478\)the expense of sugar is cheapest and when\(t = 4.6178\)the expense of sugar was expensive.
Here,\(0.85478 \approx 10\)months, that is\(10\)months from\(August{\rm{ }}1993\)is\(June{\rm{ }}1994\)and similarly,\(4.6178 \approx 4\)years 7 months, that is 4 years 7 months from\(August{\rm{ }}1993\)is\(March{\rm{ }}1998.\)
Therefore, the time when sugar was cheapest is when \(t \approx 0.855\) (\(June{\rm{ }}1994\)) and most expensive is when \(t \approx 4.618\)(March 1998) during the given period\(1993 - 2003\).