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Zorluk: OrtaLinear and Exponential Growth

At the beginning of 2015, Forest A had 4,000 trees and Forest B had 3,000 trees. The number of trees in Forest A increases by 150 trees each year. The number of trees in Forest B increases by 4% each year. If A(t)A(t) and B(t)B(t) represent the number of trees in Forest A and Forest B, respectively, tt years after the beginning of 2015, what is the value of A(5)B(5)A(5) - B(5), rounded to the nearest whole number?

Cevap: 1100 trees

Cevap

The correct value of A(5)B(5)A(5) - B(5) is 1100.
The correct value is 1100. By defining Forest A's population with the linear function A(t)=4000+150tA(t) = 4000 + 150t and Forest B's population with the exponential function B(t)=3000(1.04)tB(t) = 3000(1.04)^t, evaluating both functions at t=5t = 5 yields A(5)=4750A(5) = 4750 and B(5)3650B(5) \approx 3650. The difference is 47503650=11004750 - 3650 = 1100.

Adım Adım Çözüm

1
Formulate the growth model for Forest A.
A(t)=4000+150tA(t) = 4000 + 150t
Forest A grows linearly with a constant increase of 150 trees per year from an initial value of 4,000 trees.
2
Formulate the growth model for Forest B.
B(t)=3000(1.04)tB(t) = 3000(1.04)^t
Forest B grows exponentially with a constant percent increase of 4% per year (growth factor of 1+0.04=1.041 + 0.04 = 1.04) from an initial value of 3,000 trees.
3
Calculate the population of each forest after 5 years (t=5t = 5).
A(5)=4750A(5) = 4750 and B(5)3650B(5) \approx 3650
Substitute t=5t = 5 into both models: A(5)=4000+150(5)=4750A(5) = 4000 + 150(5) = 4750 and B(5)=3000(1.04)53649.96B(5) = 3000(1.04)^5 \approx 3649.96.
4
Calculate the difference A(5)B(5)A(5) - B(5) and round to the nearest whole number.
1100
Subtract: 47503649.96=1100.044750 - 3649.96 = 1100.04. Rounding 1100.04 to the nearest whole number yields 1100.

Anahtar Kavram

Linear vs. Exponential Growth Models

Alternatif Yöntem

Instead of evaluating each function separately, you can construct the difference expression directly as D(t)=(4000+150t)3000(1.04)tD(t) = (4000 + 150t) - 3000(1.04)^t and substitute t=5t = 5 into D(t)D(t) to compute D(5)=47503000(1.04)51100.04D(5) = 4750 - 3000(1.04)^5 \approx 1100.04, which rounds to 1100.
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