Maths help please D:

[gds08]

I've missed alot of class latley and my class is doing calculus. On this assignment all the questions are really similar so if somebody could help me answer this one then I should be right thanks .

"The rabbit population (R) on a farmer's property during a 14 months eradication program can be modelled as":
R=2(t+4) (t-14)^2 = 2t - 48t + 168t +1568
For t = 0 to 14 months.

a) initialy how many rabbits are there on the property?

b) Use calculus to determine the times at which the rabbit population on the farm is at a maximum and minimum

c) Determine the maximum number of rabbits on the farm during this time

 

[malakadang]

(a) = 1568.

(b) = The Rabbit population is at its maximum at 2 months, and minimum at 14 months.

(c) = Maximum number of rabbits @ 2 months is 1728. Minimum number of rabbits @ 14 months is 0.


Working out for (a):

Sub in 0 for t as initial time is 0, and your left over with 1568.

Working out for (b)

Find the derivative = 6t³ - 48t² +168t. Solve the derivative for 0, to find when the rate of change = 0.

0 = 6t³ - 48t² +168t; t = 2 or 14.

Working out for (c):

Sub in 2, and 14 into the original equation.

 

[Deathsnova]

I've missed alot of class latley and my class is doing calculus. On this assignment all the questions are really similar so if somebody could help me answer this one then I should be right thanks .

"The rabbit population (R) on a farmer's property during a 14 months eradication program can be modelled as":
R=2(t+4) (t-14)^2 = 2t - 48t + 168t +1568
For t = 0 to 14 months.

a) initialy how many rabbits are there on the property?

b) Use calculus to determine the times at which the rabbit population on the farm is at a maximum and minimum

c) Determine the maximum number of rabbits on the farm during this time

Double checked the first one and its correct, so im pretty sure the other 2 are right. You should be good to go OP

 

[gds08]

(a) = 1568.

(b) = The Rabbit population is at its maximum at 2 months, and minimum at 14 months.

(c) = Maximum number of rabbits @ 2 months is 1728. Minimum number of rabbits @ 14 months is 0.


Working out for (a):

Sub in 0 for t as initial time is 0, and your left over with 1568.

Working out for (b)

Find the derivative = 6t³ - 48t² +168t. Solve the derivative for 0, to find when the rate of change = 0.

0 = 6t³ - 48t² +168t; t = 2 or 14.

Working out for (c):

Sub in 2, and 14 into the original equation.

omg thanks so much. I did have the right idea in mind It just didn't make too much sense to me that theres more rabbits after 2 months when its ment to be an eradication program but I guess its right? Thanks man .

 

[malakadang]

omg thanks so much. I did have the right idea in mind It just didn't make too much sense to me that theres more rabbits after 2 months when its ment to be an eradication program but I guess its right? Thanks man .

Np!

The reason for this is because in reality, the eradication program may not take effect immediately. For example, if you spray weeds, they don't die instantly, some new ones may sprout, however, it takes a few days for the coating of the weed spray to disconnect the weeds ability to absorb water, and thus killing it.

Whenever you get a question that says where is the maximum and minimum at, always find the derivative, and solve it for 0. Whenever it says what is the maximum and or minimum, do the same thing, however, sub the value you solved in the derivative back into the original equation.

 

[xNATExDAxGR8x]

a. 1568

 

[.Liverpool.]

integrate the first one using limits, 0 to 14., double differentiate the second and you should be done.

thanks