Question

the number of hats made by a manufacturer t days after the start of the month is modeled by n(t) = 17t² + 3t, and the cost per hat, in dollars, is modeled by c(t) = ln(245t). the total cost of the hats made is v(t) = c(t)·n(t). how fast is the total cost of the hats changing on the 18th day? (round your answer to the nearest cent.)

Explanation:

Step1: Find the total cost function \( v(t) \)

We know that \( v(t)=c(t)\cdot n(t) \), where \( n(t) = 17t^{2}+3t \) and \( c(t)=\ln(245t) \). So \( v(t)=\ln(245t)\cdot(17t^{2} + 3t) \)

Step2: Use the product rule to find \( v^{\prime}(t) \)

The product rule states that if \( y = u\cdot v \), then \( y^{\prime}=u^{\prime}v+uv^{\prime} \). Let \( u=\ln(245t) \) and \( w = 17t^{2}+3t \)
- First, find \( u^{\prime}(t) \):
Using the chain rule, if \( u=\ln(245t)=\ln(245)+\ln(t) \), then \( u^{\prime}(t)=\frac{1}{t} \)
- Second, find \( w^{\prime}(t) \):
Differentiate \( w = 17t^{2}+3t \) with respect to \( t \), we get \( w^{\prime}(t)=34t + 3 \)

Now apply the product rule:
\( v^{\prime}(t)=u^{\prime}(t)w(t)+u(t)w^{\prime}(t)=\frac{1}{t}(17t^{2}+3t)+\ln(245t)(34t + 3) \)
Simplify \( \frac{1}{t}(17t^{2}+3t)=17t + 3 \)
So \( v^{\prime}(t)=17t + 3+\ln(245t)(34t + 3) \)

Step3: Evaluate \( v^{\prime}(t) \) at \( t = 18 \)

First, calculate \( 245t \) when \( t = 18 \): \( 245\times18=4410 \)
Then, calculate each part:
- \( 17t+3 \) when \( t = 18 \): \( 17\times18+3=306 + 3=309 \)
- \( \ln(4410)\approx8.39 \) (using calculator)
- \( 34t + 3 \) when \( t = 18 \): \( 34\times18+3=612+3 = 615 \)
Now, \( \ln(245t)(34t + 3)\approx8.39\times615\approx5160.85 \)
Then \( v^{\prime}(18)=309+5160.85 = 5469.85 \) (Wait, this seems wrong. Wait, let's recalculate \( \ln(245t) \) at \( t = 18 \):

\( \ln(245\times18)=\ln(4410)\approx8.3903 \)

\( 34t + 3=34\times18+3=612 + 3=615 \)

\( \ln(4410)\times615\approx8.3903\times615\approx8.3903\times600+8.3903\times15=5034.18+125.8545 = 5159.0345 \)

\( 17t+3=17\times18 + 3=306+3 = 309 \)

So \( v^{\prime}(18)=309+5159.0345=5468.0345 \)

Wait, maybe I made a mistake in the product rule. Wait, original \( c(t)=\ln(245t) \), \( n(t)=17t^{2}+3t \)

So \( v(t)=c(t)n(t)=\ln(245t)(17t^{2}+3t) \)

The derivative \( v^{\prime}(t)=\frac{d}{dt}[\ln(245t)]\times(17t^{2}+3t)+\ln(245t)\times\frac{d}{dt}(17t^{2}+3t) \)

\( \frac{d}{dt}[\ln(245t)]=\frac{245}{245t}=\frac{1}{t} \) (using chain rule: derivative of \( \ln(u) \) is \( \frac{u^{\prime}}{u} \), here \( u = 245t \), \( u^{\prime}=245 \))

\( \frac{d}{dt}(17t^{2}+3t)=34t + 3 \)

So \( v^{\prime}(t)=\frac{1}{t}(17t^{2}+3t)+\ln(245t)(34t + 3)=17t + 3+\ln(245t)(34t + 3) \)

Now plug \( t = 18 \):

\( 17\times18=306 \), so \( 17t + 3=306 + 3=309 \)

\( 245t=245\times18 = 4410 \), \( \ln(4410)\approx8.3903 \)

\( 34t + 3=34\times18+3=612 + 3=615 \)

\( \ln(4410)\times615\approx8.3903\times615 \)

Calculate \( 8\times615 = 4920 \), \( 0.3903\times615\approx0.3903\times600=234.18 \), \( 0.3903\times15 = 5.8545 \), so \( 234.18+5.8545 = 240.0345 \), total \( 4920+240.0345 = 5160.0345 \)

Then \( v^{\prime}(18)=309+5160.0345 = 5469.0345 \approx5469.03 \) (Wait, maybe my initial calculation of \( \ln(4410) \) is wrong. Let's use a calculator for \( \ln(4410) \):

\( \ln(4410)=\ln(441\times10)=\ln(21^{2}\times10)=2\ln(21)+\ln(10)\approx2\times3.0445+2.3026=6.089 + 2.3026=8.3916 \)

Then \( 8.3916\times615 \):

\( 8\times615 = 4920 \)

\( 0.3916\times615=0.3916\times600 + 0.3916\times15=234.96+5.874 = 240.834 \)

So \( 4920+240.834 = 5160.834 \)

Then \( 309+5160.834 = 5469.834 \approx5469.83 \)

Wait, maybe there is a miscalculation. Let's check the problem again. Wait, the cost per hat is \( c(t)=\ln(245t) \), number of hats is \( n(t)=17t^{2}+3t \), so total cost \( v(t)=c(t)n(t) \). We need to find the rate of change of total cost, which is \( v^{\prime}(t) \), at \( t = 18 \).

Alternative approach:

First, find \( n(18)=17\times18^{2}+3\times18=17\times324 + 54=5508+54 = 5562 \)

\( c(18)=\ln(245\times18)=\ln(4410)\approx8.3916 \)

\( n^{\prime}(t)=34t + 3 \), so \( n^{\prime}(18)=34\times18+3=615 \)

\( c^{\prime}(t)=\frac{245}{245t}=\frac{1}{t} \), so \( c^{\prime}(18)=\frac{1}{18}\approx0.0556 \)

Then by product rule \( v^{\prime}(t)=c^{\prime}(t)n(t)+c(t)n^{\prime}(t) \)

So \( v^{\prime}(18)=c^{\prime}(18)n(18)+c(18)n^{\prime}(18) \)

\( c^{\prime}(18)n(18)=\frac{1}{18}\times5562 = 309 \)

\( c(18)n^{\prime}(18)=\ln(4410)\times615\approx8.3916\times615\approx5160.83 \)

Then \( v^{\prime}(18)=309 + 5160.83=5469.83 \)

So the rate of change of total cost on the 18th day is approximately \( \$5469.83 \)

Answer:

\( 5469.83 \)