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: \( T(t)=12\times(1 + 0.4)^{t}=12\times1.4^{t}\) (since it increases by 40% each year). The number of white - tail deer \( D(t) \) follows a linear growth model: \( D(t)=18 + 10t\) (since it starts at 18 and increases by 10 each year).

Step2: Set up the inequality

We want to find the smallest \( t\geq0 \) (where \( t = 0 \) is the first year) such that \( T(t)>D(t) \), i.e., \( 12\times1.4^{t}>18 + 10t \).

Step3: Test \( t = 0 \)

For \( t = 0 \):
\( T(0)=12\times1.4^{0}=12\times1 = 12 \)
\( D(0)=18+10\times0 = 18 \)
Since \( 12<18 \), the inequality is not satisfied at \( t = 0 \).

Step4: Test \( t = 1 \)

For \( t = 1 \):
\( T(1)=12\times1.4^{1}=12\times1.4 = 16.8 \)
\( D(1)=18 + 10\times1=28 \)
Since \( 16.8<28 \), the inequality is not satisfied at \( t = 1 \).

Step5: Test \( t = 2 \)

For \( t = 2 \):
\( T(2)=12\times1.4^{2}=12\times1.96 = 23.52 \)
\( D(2)=18+10\times2 = 38 \)
Since \( 23.52<38 \), the inequality is not satisfied at \( t = 2 \).

Step6: Test \( t = 3 \)

For \( t = 3 \):
\( T(3)=12\times1.4^{3}=12\times2.744 = 32.928 \)
\( D(3)=18+10\times3 = 48 \)
Since \( 32.928<48 \), the inequality is not satisfied at \( t = 3 \).

Step7: Test \( t = 4 \)

For \( t = 4 \):
\( T(4)=12\times1.4^{4}=12\times3.8416 = 46.0992 \)
\( D(4)=18+10\times4 = 58 \)
Since \( 46.0992<58 \), the inequality is not satisfied at \( t = 4 \).

Step8: Test \( t = 5 \)

For \( t = 5 \):
\( T(5)=12\times1.4^{5}=12\times5.37824 = 64.53888 \)
\( D(5)=18+10\times5 = 68 \)
Since \( 64.53888<68 \), the inequality is not satisfied at \( t = 5 \).

Step9: Test \( t = 6 \)

For \( t = 6 \):
\( T(6)=12\times1.4^{6}=12\times7.529536 = 90.354432 \)
\( D(6)=18+10\times6 = 78 \)
Since \( 90.354432>78 \), the inequality is satisfied at \( t = 6 \). But we need to check if we made a mistake in the starting point. Wait, \( t = 0 \) is the first year. Let's re - express the years. When \( t = 0 \), it's year 1, \( t = 1 \) is year 2, \(\cdots\), \( t = 5 \) is year 6, \( t = 6 \) is year 7? Wait, no. Wait, our initial definition: when \( t = 0 \), it's the first year of counting. Let's re - evaluate the equations with the correct interpretation.

Wait, maybe a better way: Let's let \( n \) be the year number, starting at \( n = 1 \) (first year), \( n = 2 \) (second year), etc.

For year \( n \):
Number of turkeys: \( T(n)=12\times1.4^{n - 1}\) (because at \( n = 1 \), \( T(1)=12\times1.4^{0}=12 \))
Number of deer: \( D(n)=18+10\times(n - 1)\) (because at \( n = 1 \), \( D(1)=18+10\times0 = 18 \))

We want \( T(n)>D(n) \), so \( 12\times1.4^{n - 1}>18 + 10(n - 1)=8 + 10n \)

Test \( n = 1 \):
\( T(1)=12\times1.4^{0}=12 \), \( D(1)=18 \), \( 12<18 \)

Test \( n = 2 \):
\( T(2)=12\times1.4^{1}=16.8 \), \( D(2)=18 + 10\times1=28 \), \( 16.8<28 \)

Test \( n = 3 \):
\( T(3)=12\times1.4^{2}=23.52 \), \( D(3)=18+10\times2 = 38 \), \( 23.52<38 \)

Test \( n = 4 \):
\( T(4)=12\times1.4^{3}=32.928 \), \( D(4)=18+10\times3 = 48 \), \( 32.928<48 \)

Test \( n = 5 \):
\( T(5)=12\times1.4^{4}=46.0992 \), \( D(5)=18+10\times4 = 58 \), \( 46.0992<58 \)

Test \( n = 6 \):
\( T(6)=12\times1.4^{5}=64.53888 \), \( D(6)=18+10\times5 = 68 \), \( 64.53888<68 \)

Test \( n = 7 \):
\( T(7)=12\times1.4^{6}=90.354432 \), \( D(7)=18+10\times6 = 78 \), \( 90.354432>78 \)

Wait, there was a mistake in the previous variable definition. Let's do it correctly:

Let \( n \) be the year, \( n = 1,2,3,\cdots \)

At year \( n \):

Turkey population: \( T(n)=12\times(1.4)^{n - 1}\) (since in year 1, it's 12, year 2 it's \( 12\times1.4 \), etc.)

Deer population: \( D(n)=18+10\times(n - 1)\) (since in year 1, it's 18, year 2 it's \( 18 + 10 \), etc.)

We want \( T(n)>D(n) \)

\( n = 1 \): \( T(1)=12 \), \( D(1)=18 \), \( 12<18 \)

\( n = 2 \): \( T(2)=12\times1.4 = 16.8 \), \( D(2)=18 + 10=28 \), \( 16.8<28 \)

\( n = 3 \): \( T(3)=12\times1.4^{2}=12\times1.96 = 23.52 \), \( D(3)=18+20 = 38 \), \( 23.52<38 \)

\( n = 4 \): \( T(4)=12\times1.4^{3}=12\times2.744 = 32.928 \), \( D(4)=18 + 30=48 \), \( 32.928<48 \)

\( n = 5 \): \( T(5)=12\times1.4^{4}=12\times3.8416 = 46.0992 \), \( D(5)=18+40 = 58 \), \( 46.0992<58 \)

\( n = 6 \): \( T(6)=12\times1.4^{5}=12\times5.37824 = 64.53888 \), \( D(6)=18+50 = 68 \), \( 64.53888<68 \)

\( n = 7 \): \( T(7)=12\times1.4^{6}=12\times7.529536 = 90.354432 \), \( D(7)=18+60 = 78 \), \( 90.354432>78 \)

Wait, maybe we made a mistake in the deer's growth. Wait, the problem says "At the first year Sheila counted 18 white - tail deer, and their number increases by 10 additional deer per year." So in year 1: 18, year 2: \( 18 + 10=28 \), year 3: \( 28+10 = 38 \), year 4: \( 38 + 10=48 \), year 5: \( 48+10 = 58 \), year 6: \( 58+10 = 68 \), year 7: \( 68+10 = 78 \)

Turkeys: year 1:12, year 2: \( 12\times1.4 = 16.8 \), year 3: \( 16.8\times1.4=23.52 \), year 4: \( 23.52\times1.4 = 32.928 \), year 5: \( 32.928\times1.4=46.0992 \), year 6: \( 46.0992\times1.4 = 64.53888 \), year 7: \( 64.53888\times1.4=90.354432 \)

So in year 7, the number of turkeys (90.35) is more than the number of deer (78). But wait, let's check the calculation again. Wait, maybe the initial model for turkeys is wrong. Wait, the problem says "At the first year Sheila counted 12 wild turkeys, and their number increases by approximately 40% each year." So the formula for turkeys should be \( T(n)=12\times(1 + 0.4)^{n - 1}\) where \( n \) is the year number (n = 1,2,3,...). For deer, \( D(n)=18+10\times(n - 1)\)

Wait, maybe we can solve the inequality \( 12\times1.4^{n-1}>18 + 10(n - 1) \)

Simplify the right - hand side: \( 18+10n-10=8 + 10n \)

So we have \( 12\times1.4^{n - 1}-10n-8>0 \)

Let's define a function \( f(n)=12\times1.4^{n - 1}-10n - 8 \)

We can use trial and error:

\( n = 1 \): \( f(1)=12\times1-10 - 8=12 - 18=-6<0 \)

\( n = 2 \): \( f(2)=12\times1.4-20 - 8=16.8-28=-11.2<0 \) (Wait, no, \( n = 2 \): \( 12\times1.4^{1}-10\times2-8=16.8 - 20 - 8=-11.2 \))

\( n = 3 \): \( 12\times1.4^{2}-30 - 8=12\times1.96-38=23.52 - 38=-14.48<0 \)

\( n = 4 \): \( 12\times1.4^{3}-40 - 8=12\times2.744-48=32.928 - 48=-15.072<0 \)

\( n = 5 \): \( 12\times1.4^{4}-50 - 8=12\times3.8416-58=46.0992 - 58=-11.9008<0 \)

\( n = 6 \): \( 12\times1.4^{5}-60 - 8=12\times5.37824-68=64.53888 - 68=-3.46112<0 \)

\( n = 7 \): \( 12\times1.4^{6}-70 - 8=12\times7.529536-78=90.354432 - 78 = 12.354432>0 \)

So when \( n = 7 \), the inequality is satisfied. But let's check if we made a mistake in the year numbering. Wait, the first year is \( n = 1 \), so the 7th year is the first year when turkeys are more than deer. But wait, maybe there is a miscalculation. Let's try to solve the equation \( 12\times1.4^{x}=18 + 10x \) (treating \( x \) as a real number) and then find the integer value of \( x \) (year) that satisfies the inequality.

Let \( f(x)=12\times1.4^{x}-18 - 10x \)

We can use the method of successive approximation.

For \( x = 5 \): \( f(5)=12\times1.4^{5}-18 - 50=12\times5.37824-68=64.53888 - 68=-3.46112 \)

For \( x = 6 \): \( f(6)=12\times1.4^{6}-18 - 60=12\times7.529536-78=90.354432 - 78 = 12.354432 \)

Since \( f(5)<0 \) and \( f(6)>0 \), by the Intermediate Value Theorem, there is a root between \( x = 5 \) and \( x = 6 \). But since we are dealing with years (discrete values), we check the integer values.

Wait, maybe the initial problem's year - counting is different. Maybe the first year is \( t = 1 \), and we model the number of turkeys as \( T(t)=12\times1.4^{t - 1} \) and deer as \( D(t)=18+10\times(t - 1) \)

We want \( T(t)>D(t) \)

\( t = 1 \): \( T(1)=12 \), \( D(1)=18 \), no

\( t = 2 \): \( T(2)=12\times1.4 = 16.8 \), \( D(2)=28 \), no

\( t = 3 \): \( T(3)=23.52 \), \( D(3)=38 \), no

\( t = 4 \): \( T(4)=32.928 \), \( D(4)=48 \), no

\( t = 5 \): \( T(5)=46.0992 \), \( D(5)=58 \), no

\( t = 6 \): \( T(6)=64.53888 \), \( D(6)=68 \), no

\( t = 7 \): \( T(7)=90.354432 \), \( D(7)=78 \), yes

Answer:

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