Question

the following model represents the total spent on education funding, in millions of dollars, in a state from 2001 to 2011, where the time t = 0 represents the year 2001: e(t) = 28 ln(2926 − 300t) find the rate at which the total spent on education funding in that state was changing in the year 2006. (round your answer to the nearest integer.) the total spent on education funding was? by? million dollars per year in the year 2006.

Explanation:

Step1: Determine the value of \( t \) for 2006

Since \( t = 0 \) represents 2001, for 2006, \( t=2006 - 2001=5 \).

Step2: Find the derivative of \( E(t) \)

Given \( E(t)=28\ln(2926 - 300t) \), we use the chain rule. The derivative of \( \ln(u) \) is \( \frac{1}{u}u' \), where \( u = 2926-300t \) and \( u'=- 300 \).
So \( E'(t)=28\times\frac{-300}{2926 - 300t}=\frac{-8400}{2926 - 300t} \).

Step3: Substitute \( t = 5 \) into \( E'(t) \)

First, calculate the denominator when \( t = 5 \): \( 2926-300\times5=2926 - 1500 = 1426 \).
Then \( E'(5)=\frac{-8400}{1426}\approx - 5.89 \). Rounding to the nearest integer, we get \( - 6 \). The negative sign indicates a decrease, but we are asked for the rate (magnitude or considering the sign as per context, but the question says "rate at which... was changing", so we take the value after rounding. Wait, maybe I made a mistake in sign? Wait, let's re - check the derivative.
Wait, \( E(t)=28\ln(2926 - 300t) \), \( u = 2926-300t \), \( u'=-300 \), so \( E'(t)=28\times\frac{u'}{u}=28\times\frac{-300}{2926 - 300t}=\frac{-8400}{2926 - 300t} \). When \( t = 5 \), denominator is \( 2926-1500 = 1426 \), so \( E'(5)=\frac{-8400}{1426}\approx - 5.89\approx - 6 \). But maybe the question expects the absolute value? Wait, no, the rate of change can be negative (decreasing). But let's check the calculation again. Wait, maybe I messed up the function. Wait, the function is \( E(t) = 28\ln(2926-300t) \). Let's compute \( 2926-300\times5=2926 - 1500 = 1426 \). Then \( \frac{-8400}{1426}\approx - 5.89 \), which rounds to \( - 6 \). But maybe the problem has a typo, or maybe I misread the function. Wait, maybe the function is \( E(t)=28\ln(2926 + 300t) \)? If that's the case, then \( u' = 300 \), and \( E'(t)=\frac{28\times300}{2926 + 300t} \). For \( t = 5 \), denominator is \( 2926+1500 = 4426 \), \( E'(5)=\frac{8400}{4426}\approx1.897\approx2 \). But the original function is \( 2926 - 300t \). Wait, maybe the user made a typo, but based on the given function, let's proceed. Wait, maybe I miscalculated \( 2926-300\times5 \). \( 300\times5 = 1500 \), \( 2926-1500=1426 \). Then \( 8400\div1426\approx5.89 \). So the rate of change is approximately - 6 million dollars per year. But maybe the question wants the absolute value, but the derivative gives the rate (including sign). However, let's check again.

Wait, maybe the function is \( E(t)=28\ln(2926 + 300t) \). Let's assume that maybe there was a sign error in the problem. If \( E(t)=28\ln(2926 + 300t) \), then \( E'(t)=28\times\frac{300}{2926 + 300t}=\frac{8400}{2926+300t} \). For \( t = 5 \), \( 2926 + 1500=4426 \), \( 8400\div4426\approx1.897\approx2 \). But since the original function is \( 2926 - 300t \), we have to go with that. Wait, maybe the problem is about the rate of change, and we just need to compute the value. Let's do the calculation again:

\( E'(t)=\frac{d}{dt}[28\ln(2926 - 300t)]=28\times\frac{- 300}{2926-300t}=\frac{-8400}{2926 - 300t} \)

At \( t = 5 \):

\( 2926-300\times5=2926 - 1500 = 1426 \)

\( E'(5)=\frac{-8400}{1426}\approx - 5.89\approx - 6 \)

So the rate at which the total spent on education funding was changing in 2006 is approximately - 6 million dollars per year. If we consider the magnitude (the rate of change in terms of amount, ignoring the sign of decrease), but the derivative gives the rate (with sign indicating decrease). However, maybe the problem expects the absolute value, but based on the calculation, the answer is - 6 (or 6 if we consider the magnitude of the rate of change). Wait, maybe I made a mistake in the derivative. Let's recall the derivative of \( \ln(u) \) is \( \frac{u'}{u} \), so if \( u = 2926-300t \), \( u'=-300 \), so \( E'(t)=28\times\frac{-300}{2926 - 300t} \), that's correct.

Answer:

\(-6\) (or \(6\) if we consider the magnitude of the rate of change, but based on the derivative calculation, it's \(-6\))