Question

sheila is a wildlife biologist. at the beginning of each year, she tracks the wild turkey and white - tail deer population in the game reserve where she works. - at the first year sheila counted 12 wild turkeys, and their number increases by approximately 40% each year. - at the first year sheila counted 18 white - tail deer, and their number increases by 10 additional deer per year. what is the first year in which sheila counts more turkeys than deer?

Explanation:

Step1: Define the population functions

Let \( t \) be the number of years after the first year. The number of wild turkeys \( T(t) \) follows an exponential growth model: \( T(t)=12(1 + 0.4)^t=12(1.4)^t \). The number of white - tail deer \( D(t) \) follows a linear growth model: \( D(t)=18 + 10t \).

Step2: Set up the inequality

We want to find the smallest \( t\) (where \( t\) is a non - negative integer) such that \( T(t)>D(t) \), i.e., \( 12(1.4)^t>18 + 10t \).

Step3: Test values of \( t\)

• For \( t = 0\):

\( T(0)=12(1.4)^0 = 12\), \( D(0)=18+10(0)=18 \). Since \( 12<18 \), the inequality is not satisfied.

• For \( t = 1\):

\( T(1)=12\times1.4=16.8\), \( D(1)=18 + 10\times1=28 \). Since \( 16.8<28 \), the inequality is not satisfied.

• For \( t = 2\):

\( T(2)=12\times(1.4)^2=12\times1.96 = 23.52\), \( D(2)=18+10\times2 = 38 \). Since \( 23.52<38 \), the inequality is not satisfied.

• For \( t = 3\):

\( T(3)=12\times(1.4)^3=12\times2.744 = 32.928\), \( D(3)=18+10\times3=48 \). Since \( 32.928<48 \), the inequality is not satisfied.

• For \( t = 4\):

\( T(4)=12\times(1.4)^4=12\times3.8416 = 46.0992\), \( D(4)=18+10\times4 = 58 \). Since \( 46.0992<58 \), the inequality is not satisfied.

• For \( t = 5\):

\( T(5)=12\times(1.4)^5=12\times5.37824 = 64.53888\), \( D(5)=18+10\times5=68 \). Since \( 64.53888<68 \), the inequality is not satisfied.

• For \( t = 6\):

\( T(6)=12\times(1.4)^6=12\times7.529536 = 90.354432\), \( D(6)=18+10\times6 = 78 \). Since \( 90.354432>78 \), the inequality is satisfied.

Since \( t = 6\) corresponds to the \( 6 + 1=7\)th year (because \( t = 0\) is the first year of counting), we need to check if we made a mistake in the index. Wait, actually, when \( t = 0\), it's the first year of counting. Let's re - index: Let \( n=t + 1\) be the year number. When \( t=5\) (year \( n = 6\)):
Wait, no, let's re - evaluate the starting point. The first year of counting is when \( t = 0\) (initial count). Let's re - check the calculation for \( t=5\):
\( (1.4)^5=1.4\times1.4\times1.4\times1.4\times1.4=5.37824\), \( 12\times5.37824 = 64.53888\), \( D(5)=18 + 50=68\), \( 64.53888<68\)
For \( t = 6\):
\( (1.4)^6=1.4\times5.37824 = 7.529536\), \( 12\times7.529536=90.354432\), \( D(6)=18+60 = 78\), \( 90.354432>78\)

Since \( t = 6\) means 6 years after the first count, the year number is \( 6 + 1=7\)? Wait, no. Wait, the first count is at \( t = 0\) (year 1), \( t = 1\) is year 2, \( t=2\) is year 3, \( t = 3\) is year 4, \( t=4\) is year 5, \( t = 5\) is year 6, \( t=6\) is year 7? Wait, no, let's start over with the year definition. Let the first year (initial count) be year \( y = 1\). Then for year \( y\), \( t=y - 1\).

We can also set up the equation with \( y\) as the year number, where \( y\geq1\). Then \( T(y - 1)=12(1.4)^{y - 1}\) and \( D(y - 1)=18+10(y - 1)=8 + 10y\).

We want \( 12(1.4)^{y - 1}>8 + 10y\)

For \( y = 1\): \( 12(1.4)^0=12\), \( 8 + 10(1)=18\), \( 12<18\)

For \( y = 2\): \( 12(1.4)^1 = 16.8\), \( 8+10(2)=28\), \( 16.8<28\)

For \( y = 3\): \( 12(1.4)^2=23.52\), \( 8 + 10(3)=38\), \( 23.52<38\)

For \( y = 4\): \( 12(1.4)^3=32.928\), \( 8+10(4)=48\), \( 32.928<48\)

For \( y = 5\): \( 12(1.4)^4=46.0992\), \( 8 + 10(5)=58\), \( 46.0992<58\)

For \( y = 6\): \( 12(1.4)^5=64.53888\), \( 8+10(6)=68\), \( 64.53888<68\)

For \( y = 7\): \( 12(1.4)^6=90.354432\), \( 8+10(7)=78\), \( 90.354432>78\)

Answer:

The first year in which Sheila counts more turkeys than deer is the 7th year.