Its the old rabbit colony question using simultaneous linear and quadratic equations - and I need help:
A colony of rabbits is growing so that the total number of rabbits (R) after number of weeks (t) is shown: on this table.
t=5 R=140, t=10 R=200, t=15 R=280, t=20 R=400, t=25 R=570, t=30 R=800
The food supply available is such that it can feed (N) of rabbits at any given time (t = weeks) according to the table:
t=5 N=440, t=10 R=480, t=15 R=520, t=20 R=560, t=25 R=600.
At what point in time is the population of rabbits and food balances?
I know I have to use simultaneous linear and quadratic equations - I just need to put them into equations first.
Thanks for the help.
To solve this problem, we can set up two equations - one represents the population of rabbits (R) over time (t), and the other represents the food supply (N) over time (t).
Let's assume the equation for the population of rabbits is a quadratic equation of the form:
R = at^2 + bt + c
To find the values of a, b, and c, we can plug in the given points (t, R) from the table into the equation.
Using the given points (t=5, R=140), (t=10, R=200), and (t=15, R=280), we can set up the following equations:
140 = 5a(5^2) + 5b + c
200 = 10a(10^2) + 10b + c
280 = 15a(15^2) + 15b + c
Simplifying each equation gives:
140 = 25a + 5b + c ...(equation 1)
200 = 100a + 10b + c ...(equation 2)
280 = 225a + 15b + c ...(equation 3)
Similarly, we assume the equation for the food supply is a linear equation of the form:
N = pt + q
Using the given points (t=5, N=440) and (t=10, N=480), we can set up the following equations:
440 = 5p + q
480 = 10p + q
Now, we have a system of four equations:
140 = 25a + 5b + c ...(equation 1)
200 = 100a + 10b + c ...(equation 2)
280 = 225a + 15b + c ...(equation 3)
440 = 5p + q ...(equation 4)
480 = 10p + q ...(equation 5)
To solve this system, we can use any method, such as substitution or elimination. Let's use substitution:
From equation 4, we can solve for q:
q = 440 - 5p
Substitute this value of q into equation 5:
480 = 10p + (440 - 5p)
480 = 10p + 440 - 5p
480 = 5p + 440
5p = 40
p = 8
Now we can substitute the value of p back into equation 4 to find q:
440 = 5(8) + q
440 = 40 + q
q = 400
So, the equation for the food supply is:
N = 8t + 400
Next, we can substitute the values of a, b, and c into the quadratic equation for the rabbit population (equations 1-3):
140 = 25a + 5b + c ...(equation 1)
200 = 100a + 10b + c ...(equation 2)
280 = 225a + 15b + c ...(equation 3)
By solving these three equations simultaneously, we can determine the values of a, b, and c. Once we have these values, we can find at what point in time the population of rabbits and food supply will be balanced by setting the two equations equal to each other:
at^2 + bt + c = 8t + 400
Solving this equation will give us the point in time when the population of rabbits and food supply are balanced.