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mattie_j37890's question at Yahoo! Answers regarding approximating critical values

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MarkFL

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Feb 24, 2012
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Here is the question:

What are the critical values of this function?

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 − 0.04467t + 0.4438
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.)
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Re: mattie_j37890's quetion at Yahoo! Answers regarding approximating critical values

Hello mattie_j37890,

We are given the function:

\(\displaystyle S(t)=-0.00003237t^5+0.0009037t^4-0.008956t^3+0.03629t^2-0.04467t+0.4438\)

First, let's look at a plot of this function just so we have some idea of what it looks like over the relevant domain:

mattie1.jpg

We can see that there are 4 relative extrema in the interval $0\le t\le10$

To find where they are, that is for what values of $t$ do they exist, we want to equate the first derivative to zero:

\(\displaystyle S'(t)=-0.00016185t^4+0.0036148t^3-0.026868t^2+0.07258t-0.04467=0\)

Let's look at a plot of the derivative to get some idea of where the roots are:

mattie2.jpg

We may use Newton's method to approximate these roots. We may write:

\(\displaystyle f(t)=-0.00016185t^4+0.0036148t^3-0.026868t^2+0.07258t-0.04467=0\)

\(\displaystyle f'(t)=-0.0006474t^3+0.0108444t^2-0.053736t+0.07258\)

And so we have the recursion:

\(\displaystyle t_{n+1}=t_{n}-\frac{f\left(t_{n} \right)}{f'\left(t_{n} \right)}\)

We see the first root is near $t=1$, and so we may compute:

\(\displaystyle t_0=1\)

\(\displaystyle t_1\approx0.845220550257\)

\(\displaystyle t_2\approx0.857337077417\)

\(\displaystyle t_3\approx0.857416179268\)

\(\displaystyle t_4\approx0.857416182626\)

\(\displaystyle t_5\approx0.857416182626\)

We see the second root is near $t=5$, and so we may compute:

\(\displaystyle t_0=5\)

\(\displaystyle t_1\approx4.53064243449\)

\(\displaystyle t_2\approx4.60497652706\)

\(\displaystyle t_3\approx4.60659458701\)

\(\displaystyle t_4\approx4.606595386\)

\(\displaystyle t_5\approx4.606595386\)

We see the third root is near $t=7$, and so we may compute:

\(\displaystyle t_0=7\)

\(\displaystyle t_1\approx7.3250339402\)

\(\displaystyle t_2\approx7.30609727792\)

\(\displaystyle t_3\approx7.30607003007\)

\(\displaystyle t_4\approx7.30607003001\)

\(\displaystyle t_5\approx7.30607003001\)

We see the fourth root is near $t=10$, and so we may compute:

\(\displaystyle t_0=10\)

\(\displaystyle t_1\approx9.66222062004\)

\(\displaystyle t_2\approx9.57077528295\)

\(\displaystyle t_3\approx9.56421133076\)

\(\displaystyle t_4\approx9.56417851958\)

\(\displaystyle t_5\approx9.56417851876\)

\(\displaystyle t_6\approx9.56417851876\)

And so, we have the four roots of the first derivative approximated by:

\(\displaystyle t\approx0.857416182626,\,4.606595386,\,7.30607003001,\,9.56417851876\)

Now, to find the absolute extrema on the given closed interval, we need to evaluate the function at the endpoints and at the critical values:

\(\displaystyle S(0)=0.4438\)

\(\displaystyle S(0.857416182626)=0.4270063563925\)

\(\displaystyle S(4.606595386)=0.47243084424924303\)

\(\displaystyle S(7.30607003001)=0.462862495347419\)

\(\displaystyle S(9.56417851876)=0.47195253062506465\)

\(\displaystyle S(10)=0.4701\)

And so rounding to 3 decimal places, we find:

The absolute minimum occurs when \(\displaystyle t\approx0.857\)

The absolute maximum occurs when \(\displaystyle t\approx4.607\)
 
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