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Sagot :
As per the given question the median size of a new house and the values of different parts of the question will be as follows:
(a) We must locate the period when the rate of change is the smallest in order to establish the period during which the median house size increased the least quickly. Finding the derivative of the following equation will provide us with the rate of change of the house size with respect to time, which we can utilize to do this.
derivative of the function H(x) = 0.359x3 - 15.198x2 + 221.738x + 862.514 given by the equation for the rate of change of the house size with respect to time as
H'(x) = 3 (0.359) x2 - 2 (15.198) x + 221.738
(b) By setting the derivative to 0 and figuring out x, we may determine the moment at which the rate of change is the least. We have x = 3.904 as a result. The time where the rate of change is the smallest is 3.904 years after 1980, or 1984.904, because x denotes the number of years after 1980. This roughly corresponds to 1984, making 1984 the year in which the median house size increased at the slowest rate.
We may enter the value of x into the previous equation to obtain H(1984) = 0.359 x [tex](1984)^{3}[/tex] - 15.198 x [tex](1984)^{2}[/tex] + 221.738 ( 984) + 862.514
= 2,532,358 sq ft. as the comparable house size at this time. In 1984, this was the size of the typical home.
We may enter the value of x into the derivative equation to obtain H'(1984) = 3 (0.359) x [tex](1984)^{2}[/tex] - 2 (15.198 x 1984 + 221.738) = 606.696 sq ft per year, which corresponds to the rate of growth in house size. This represents the median house size's rate of change during the slowest period of growth.
(c) The biggest rate of change must be found in order to pinpoint the period in which the median house size increased most quickly. Finding the derivative equation's greatest value will help us do this. To accomplish this, one can solve for x while setting the derivative equal to 0, then determine the values of x that are either more or less than the answers we obtained. The derivative equation can then be used to solve for the maximum value using these x values.
Using the quadratic formula to solve for x. The quadratic formula is given by x = [tex]\frac{(-b ± √(b^2 - 4ac))}{2a}[/tex], where a, b, and c are the coefficients of the quadratic equation. Here, a = 3 (0.359) = 1.077, b = -2 (15.198) = -30.396, and c = 221.738. Putting these values into the quadratic formula gives us x = [tex]\frac{(30.396 ± √(30.396^2 - 4 (1.077 X 221.738))}{2 (1.077)}[/tex]
= [tex]\frac{(30.396 ± √(928.822 - 948.512))}{2.154}[/tex]
= [tex]\frac{(30.396 ± √(-19.690))}{2.154}[/tex]
= [tex]\frac{15.198}{2.154}[/tex]
= 7.096.
The derivative equation will have a maximum or lowest value at the value of x that we just discovered because it is a quadratic. We can enter this value of x into the derivative equation to obtain H'(7.096) = 3 x 0 and use this result to determine the derivative equation's maximum value.
To learn more about median size: https://brainly.com/question/2292760
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