Question

a large company determines that in t months, the number of units (in the thousands) that they will have produced is given by $p(t)=\frac{(11t + 1)^4}{3t^4+3}$. in the long run, what is the expected number of units produced?

Explanation:

Step1: Find the limit as t approaches infinity

We need to find $\lim_{t
ightarrow\infty}p(t)=\lim_{t
ightarrow\infty}\frac{(11t + 1)^4}{3t^4+3}$.

Step2: Expand the numerator

Using the binomial expansion $(a + b)^4=a^4+4a^3b + 6a^2b^2+4ab^3 + b^4$, for $a = 11t$ and $b = 1$, we have $(11t+1)^4=(11t)^4+4(11t)^3\times1+6(11t)^2\times1^2+4(11t)\times1^3+1^4=14641t^4 + 5324t^3+726t^2 + 44t+1$.

Step3: Divide numerator and denominator by $t^4$

$\lim_{t
ightarrow\infty}\frac{(11t + 1)^4}{3t^4+3}=\lim_{t
ightarrow\infty}\frac{14641+\frac{5324}{t}+\frac{726}{t^2}+\frac{44}{t^3}+\frac{1}{t^4}}{3+\frac{3}{t^4}}$.

Step4: Evaluate the limit

As $t
ightarrow\infty$, $\frac{1}{t},\frac{1}{t^2},\frac{1}{t^3},\frac{1}{t^4}
ightarrow0$. So $\lim_{t
ightarrow\infty}\frac{14641+\frac{5324}{t}+\frac{726}{t^2}+\frac{44}{t^3}+\frac{1}{t^4}}{3+\frac{3}{t^4}}=\frac{14641}{3}=\frac{14641}{3}\approx4880.33$.

Answer:

$\frac{14641}{3}\approx4880.33$