Respuesta :
Answer:
[tex]t = 0.811\,s[/tex] contains the cheapest reference to sugar; [tex]t = 4.511\,s[/tex] contains the most expensive reference to sugar.
Step-by-step explanation:
Let be [tex]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[/tex], 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
[tex]s'(t) = -0.00016185\cdot t^{4}+0.0036148\cdot t^{3}-0.026868\cdot t^{2}+0.07258\cdot t - 0.04547[/tex]
Let equalize the polynomial to zero and solve the resulting expression:
[tex]-0.00016185\cdot t^{4}+0.0036148\cdot t^{3}-0.026868\cdot t^{2}+0.07258\cdot t - 0.04547 = 0[/tex]
[tex]t_{1} \approx 9.511\,s[/tex], [tex]t_{2}\approx 7.431\,s[/tex], [tex]t_{3}\approx 4.511\,s[/tex] and [tex]t_{4}\approx 0.881\,s[/tex]
Second Derivative Test
[tex]s''(t) = -0.0006474\cdot t^{3}+0.0108444\cdot t^{2}-0.053736\cdot t+0.07258[/tex]
This function is now evaluated at each root found in the First Derivative section:
[tex]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[/tex]
[tex]s''(9.511\,s) = -0.015[/tex] (A maximum)
[tex]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[/tex]
[tex]s''(7.431\,s) = 6.440\times 10^{-3}[/tex] (A minimum)
[tex]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[/tex]
[tex]s''(4.511\,s) = -8.577\times 10^{-3}[/tex] (A maximum)
[tex]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[/tex]
[tex]s''(0.811\,s) = 0.036[/tex] (A minimum)
Each value is evaluated in order to determine when sugar was the cheapest and the most expensive:
Cheapest (Absolute minimum)
[tex]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[/tex]
[tex]s(0.811\,s) = 0.460[/tex]
[tex]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[/tex]
[tex]s(7.431\,s) = 0.491[/tex]
[tex]t = 0.811\,s[/tex] contains the cheapest reference to sugar.
Most expensive (Absolute maximum)
[tex]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[/tex]
[tex]s(4.511\,s) = 0.503[/tex]
[tex]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[/tex]
[tex]s(9.511\,s) = 0.498[/tex]
[tex]t = 4.511\,s[/tex] contains the most expensive reference to sugar.
The required values are,
[tex]t=0.881199[/tex] at the cheapest.
[tex]t=4.51081[/tex] at the most expensive.
Minimum or Maximum:
A high point is called a maximum (plural maxima ). A low point is called a minimum (plural minima ).
Given equation is,
[tex]S(t) = -0.00003237t^5 + 0.0009037t^4- 0.008956t^3 + 0.03629t^2-0.04547t + 0.4778[/tex]
Differentiating the given equation we get,
[tex]S'(t)=-0.00003237\times 5t^4+0.0009037\times 4t^3-0.008956\times 3t^2+0.03629\times 2t-0.04547+0\\S'(t)=0\\-0.00003237\times 5t^4+0.0009037\times 4t^3-0.008956\times 3t^2+0.03629\times 2t-0.04547+0=0\\t=0.881199\\t=4.51081\\t=7.43087\\t=9.51137\\[/tex]
Now we can directly plug those fours values of t into given function S(t) to find which one gives max or minimum or you can also use the 2nd derivative test. Although that is not compulsory
[tex]t=0.881199,S(t)=0.46031095\\t=4.51081, S(t)=0.50278423\\t=7.43087, S(t)=0.49096762\\t=9.51137, S(t)=0.49832202\\[/tex]
We see that sugar is cheapest at [tex]t=0.881199[/tex] which is approx 1 and corresponds to the year [tex]1993+1=1994[/tex]
Similarly sugar is most expensive at [tex]t=4.51081[/tex] which is approx 5 and corresponds to year [tex]1993+5=1998[/tex]
Learn more about the topic Minimum or Maximum:
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