Question
A model for the food-price index (the price of a representative "basket" of foods) between 1984 and 1994 is given by the function
I(t)=0.00009045t5+0.001438t4−0.06561t3+0.4598t2−.6270t+99.33
Where t is measured in years since midyear 1984, so 0≤t≤10, and I(t) is measured in 1987 dollars and scaled such that I(3)=100. Estimate to two decimal places the times when food was cheapest and most expensive during the 1984-1994 period.
I(t)=0.00009045t5+0.001438t4−0.06561t3+0.4598t2−.6270t+99.33
Where t is measured in years since midyear 1984, so 0≤t≤10, and I(t) is measured in 1987 dollars and scaled such that I(3)=100. Estimate to two decimal places the times when food was cheapest and most expensive during the 1984-1994 period.
Answers
Given a polynomial function in t where
I(t) = 0.00009045t5+0.001438t4−0.06561t3+0.4598t2−.6270t+99.33
We need to find the extrema for 0≤t≤10.
Steps:
find first derivative I'(0), equate to zero. Solve for the zeroes which correspond to the extrema, say t1 and t2.
To distinguish minima from maxima, evaluate I"(t1) and I"(t2) (second derivative).
If I"(t)<0, t is a maximum (most expensive), if I"(t)>0, t is a minimum (cheapest).
Hint:
t1 and t2 are near 1 and 5 resectively, in case you need to solve for zeroes using Newton's method.
I(t) = 0.00009045t5+0.001438t4−0.06561t3+0.4598t2−.6270t+99.33
We need to find the extrema for 0≤t≤10.
Steps:
find first derivative I'(0), equate to zero. Solve for the zeroes which correspond to the extrema, say t1 and t2.
To distinguish minima from maxima, evaluate I"(t1) and I"(t2) (second derivative).
If I"(t)<0, t is a maximum (most expensive), if I"(t)>0, t is a minimum (cheapest).
Hint:
t1 and t2 are near 1 and 5 resectively, in case you need to solve for zeroes using Newton's method.
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