Answer:
N(1)=50 is a minimum
N(15)=4391.7 is a maximum
Step-by-step explanation:
<u>Extrema values of functions
</u>
If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.
We are given a function (corrected)
![N(t) = 20(t^2-lnt^2)+ 30](https://tex.z-dn.net/?f=N%28t%29%20%3D%2020%28t%5E2-lnt%5E2%29%2B%2030)
![N(t) = 20(t^2-2lnt)+ 30](https://tex.z-dn.net/?f=N%28t%29%20%3D%2020%28t%5E2-2lnt%29%2B%2030)
(a)
First, we take its derivative
![N'(t) = 20(2t-\frac{2}{t})](https://tex.z-dn.net/?f=N%27%28t%29%20%3D%2020%282t-%5Cfrac%7B2%7D%7Bt%7D%29)
Solve N'(t)=0
![20(2t-\frac{2}{t})=0](https://tex.z-dn.net/?f=20%282t-%5Cfrac%7B2%7D%7Bt%7D%29%3D0)
Simplifying
![2t^2-2=0](https://tex.z-dn.net/?f=2t%5E2-2%3D0)
Solving for t
![t=1\ ,t=-1](https://tex.z-dn.net/?f=t%3D1%5C%20%2Ct%3D-1)
Only t=1 belongs to the valid interval ![1\leqslant t\leqslant 15](https://tex.z-dn.net/?f=1%5Cleqslant%20t%5Cleqslant%2015)
Taking the second derivative
![N''(t) = 20(2+\frac{2}{t^2})](https://tex.z-dn.net/?f=N%27%27%28t%29%20%3D%2020%282%2B%5Cfrac%7B2%7D%7Bt%5E2%7D%29)
Which is always positive, so t=1 is a minimum
(b)
![N(1)=20(1^2-2ln1)+ 30](https://tex.z-dn.net/?f=N%281%29%3D20%281%5E2-2ln1%29%2B%2030)
N(1)=50 is a minimum
(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15
(d)
![N(15)=20(15^2-2ln15)+ 30](https://tex.z-dn.net/?f=N%2815%29%3D20%2815%5E2-2ln15%29%2B%2030)
N(15)=4391.7 is a maximum