Question

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) = −0.00003237t5 + 0.0009037t4 − 0.008956t3 + 0.03629t2 − 0.04547t + 0.4778

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. (Round your answers to three decimal places.)

t= __________________________ (cheapest)

t=__________________________ (most expensive)

Answer 1

Answer:

$t = 0.811\,s$ contains the cheapest reference to sugar; $t = 4.511\,s$ contains the most expensive reference to sugar.

Step-by-step explanation:

Let be $s(t) = -0.00003237\cdot t^{5} + 0.0009037\cdot t^{4}-0.008956\cdot t^{3}+0.03629\cdot t^{2}-0.04547\cdot t + 0.4778$, the times when sugar is the cheapest and the most expensive (absolute minimum and maximum) are determined with the help of first and second derivatives of this function (First and Second Derivative Tests):

First Derivative Test

$s'(t) = -0.00016185\cdot t^{4}+0.0036148\cdot t^{3}-0.026868\cdot t^{2}+0.07258\cdot t - 0.04547$

Let equalize the polynomial to zero and solve the resulting expression:

$-0.00016185\cdot t^{4}+0.0036148\cdot t^{3}-0.026868\cdot t^{2}+0.07258\cdot t - 0.04547 = 0$

$t_{1} \approx 9.511\,s$, $t_{2}\approx 7.431\,s$, $t_{3}\approx 4.511\,s$ and $t_{4}\approx 0.881\,s$

Second Derivative Test

$s''(t) = -0.0006474\cdot t^{3}+0.0108444\cdot t^{2}-0.053736\cdot t+0.07258$

This function is now evaluated at each root found in the First Derivative section:

$s''(9.511\,s) = -0.0006474\cdot (9.511\,s)^{3}+0.0108444\cdot (9.511\,s)^{2}-0.053736\cdot (9.511\,s)+0.07258$

$s''(9.511\,s) = -0.015$ (A maximum)

$s''(7.431\,s) = -0.0006474\cdot (7.431\,s)^{3}+0.0108444\cdot (7.431\,s)^{2}-0.053736\cdot (7.431\,s)+0.07258$

$s''(7.431\,s) = 6.440\times 10^{-3}$ (A minimum)

$s''(4.511\,s) = -0.0006474\cdot (4.511\,s)^{3}+0.0108444\cdot (4.511\,s)^{2}-0.053736\cdot (4.511\,s)+0.07258$

$s''(4.511\,s) = -8.577\times 10^{-3}$ (A maximum)

$s''(0.811\,s) = -0.0006474\cdot (0.811\,s)^{3}+0.0108444\cdot (0.811\,s)^{2}-0.053736\cdot (0.811\,s)+0.07258$

$s''(0.811\,s) = 0.036$ (A minimum)

Each value is evaluated in order to determine when sugar was the cheapest and the most expensive:

Cheapest (Absolute minimum)

$s(0.811\,s) = -0.00003237\cdot (0.811\,s)^{5}+0.0009037\cdot (0.811\,s)^{4}-0.008956\cdot (0.811\,s)^{3}+0.03629\cdot (0.811\,s)^{2}-0.04547\cdot (0.811\,s)+0.4778$

$s(0.811\,s) = 0.460$

$s(7.431\,s) = -0.00003237\cdot (7.431\,s)^{5}+0.0009037\cdot (7.431\,s)^{4}-0.008956\cdot (7.431\,s)^{3}+0.03629\cdot (7.431\,s)^{2}-0.04547\cdot (7.431\,s)+0.4778$

$s(7.431\,s) = 0.491$

$t = 0.811\,s$ contains the cheapest reference to sugar.

Most expensive (Absolute maximum)

$s(4.511\,s) = -0.00003237\cdot (4.511\,s)^{5}+0.0009037\cdot (4.511\,s)^{4}-0.008956\cdot (4.511\,s)^{3}+0.03629\cdot (4.511\,s)^{2}-0.04547\cdot (4.511\,s)+0.4778$

$s(4.511\,s) = 0.503$

$s(9.511\,s) = -0.00003237\cdot (9.511\,s)^{5}+0.0009037\cdot (9.511\,s)^{4}-0.008956\cdot (9.511\,s)^{3}+0.03629\cdot (9.511\,s)^{2}-0.04547\cdot (9.511\,s)+0.4778$

$s(9.511\,s) = 0.498$

$t = 4.511\,s$ contains the most expensive reference to sugar.