Answer:
Explanation:
Given:
The equation describing the forest wood biomass per hectare as a function of plantation age t is:
y(t) = 5 + 0.005t^2 + 0.024t^3 − 0.0045t^4
The equation that describes the annual growth in wood biomass is:
y ′ (t) = 0.01t + 0.072t^2 - 0.018t^3
To find:
a) The year the annual growth achieved its highest possible value
b) when does y ′ (t) achieve its highest value?
a)
To determine the year the highest possible value was achieved, we will set the derivative y'(t) to zero. The values of t will be substituted into the second derivative to get the highest value
![\begin{gathered} 0\text{ = 0.01t + 0.072t}^2\text{ - 0.018t}^3 \\ using\text{ a graphing tool,} \\ t\text{ = -0.1343} \\ \text{t = 0} \\ t\text{ = 4.1343} \\ \\ Since\text{ we can't have time as a negative value, t = 0 or t = 4.1343} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%200%5Ctext%7B%20%3D%200.01t%20%2B%200.072t%7D%5E2%5Ctext%7B%20-%200.018t%7D%5E3%20%5C%5C%20using%5Ctext%7B%20a%20graphing%20tool%2C%7D%20%5C%5C%20t%5Ctext%7B%20%20%3D%20-0.1343%7D%20%5C%5C%20%5Ctext%7Bt%20%3D%200%7D%20%5C%5C%20t%5Ctext%7B%20%3D%204.1343%7D%20%5C%5C%20%20%5C%5C%20Since%5Ctext%7B%20we%20can%27t%20have%20time%20as%20a%20negative%20value%2C%20t%20%3D%200%20or%20t%20%3D%204.1343%7D%20%5Cend%7Bgathered%7D)
![\begin{gathered} To\text{ ascertain which is the maximum, we take a second derivative} \\ y^{\prime}\text{'\lparen t\rparen = 0.01 + 0.144t - 0.054t}^2 \\ \\ substitute\text{ t = 0 and t = 4.13 respectively} \\ when\text{ t = 0 } \\ y^{\prime}\text{'\lparen t\rparen= 0.01 + 0.144\lparen0\rparen - 0.054\lparen0\rparen}^2 \\ y^{\prime}\text{'\lparen t\rparen= 0.01 + 0 - 0} \\ y^{\prime}\text{'\lparen t\rparen= 0.01 > 0} \\ \\ y^{\prime}\text{'\lparen t\rparen= 0.01 + 0.144\lparen4.13\rparen - 0.054\lparen4.13\rparen}^2 \\ y^{\prime}\text{'\lparen t\rparen= -0.316 < 0} \\ \\ if\text{ }y^{\prime}\text{'\lparen t\rparen > 0 , then it is minimum, } \\ if\text{ }y^{\prime}\text{'\lparen t\rparen < 0, then it is maximum} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20To%5Ctext%7B%20ascertain%20which%20is%20the%20maximum%2C%20we%20take%20a%20second%20derivative%7D%20%5C%5C%20y%5E%7B%5Cprime%7D%5Ctext%7B%27%5Clparen%20t%5Crparen%20%3D%200.01%20%2B%200.144t%20-%200.054t%7D%5E2%20%5C%5C%20%20%5C%5C%20substitute%5Ctext%7B%20t%20%3D%200%20and%20t%20%3D%204.13%20respectively%7D%20%5C%5C%20when%5Ctext%7B%20t%20%3D%200%20%7D%20%5C%5C%20y%5E%7B%5Cprime%7D%5Ctext%7B%27%5Clparen%20t%5Crparen%3D%200.01%20%2B%200.144%5Clparen0%5Crparen%20-%200.054%5Clparen0%5Crparen%7D%5E2%20%5C%5C%20y%5E%7B%5Cprime%7D%5Ctext%7B%27%5Clparen%20t%5Crparen%3D%200.01%20%2B%200%20-%200%7D%20%5C%5C%20y%5E%7B%5Cprime%7D%5Ctext%7B%27%5Clparen%20t%5Crparen%3D%200.01%20%3E%200%7D%20%5C%5C%20%20%5C%5C%20y%5E%7B%5Cprime%7D%5Ctext%7B%27%5Clparen%20t%5Crparen%3D%200.01%20%2B%200.144%5Clparen4.13%5Crparen%20-%200.054%5Clparen4.13%5Crparen%7D%5E2%20%5C%5C%20y%5E%7B%5Cprime%7D%5Ctext%7B%27%5Clparen%20t%5Crparen%3D%20-0.316%20%3C%200%7D%20%5C%5C%20%20%5C%5C%20if%5Ctext%7B%20%7Dy%5E%7B%5Cprime%7D%5Ctext%7B%27%5Clparen%20t%5Crparen%20%3E%200%20%2C%20then%20it%20is%20minimum%2C%20%7D%20%5C%5C%20if%5Ctext%7B%20%7Dy%5E%7B%5Cprime%7D%5Ctext%7B%27%5Clparen%20t%5Crparen%20%3C%200%2C%20then%20it%20is%20maximum%7D%20%5Cend%7Bgathered%7D)
SInce t = 4.13, gives y ′' (t) = -0.316 (< 0). This makes it the maximum value of t
The year the annual growth achieved its highest possible value to the nearest whole number will be
year 4
b) y ′ (t) will achieve its highest value, when we substitute the value of t that gives into the initial function.
Initial function: y(t) = 5 + 0.005t^2 + 0.024t^3 − 0.0045t^4
Answer:
You have not added the options, therefore, I cannot provide an exact answer. However, I can help you with the concept.
In any fraction, the denominator cannot be equal to zero because this would make the fraction undefined.
Excluded values are the values that would make the denominator equal to zero.
We are given that the excluded values are 2 and 5, this means that the factors of the denominator are (x-2) and (x-5)
This means that the denominator is x² - 7x + 10
Pick the fraction which has this denominator.
Hope this helps :)
I think r should equal 32/7
74.3 is greater than 7.43
First replace x and y by 3 and 4.
(3*(x)+1)/(4*(y)^2)
((3*(3))+1)/(4*((4)^2)) = (9+1)/(4*16) = 10/64 = 5/32.
5/32 is your answer.
Please mark me as brainliest!