Answer :
Answer:
N(1)=50 is a minimum
N(15)=4391.7 is a maximum
Step-by-step explanation:
Extrema values of functions
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)
[tex]N(t) = 20(t^2-lnt^2)+ 30[/tex]
[tex]N(t) = 20(t^2-2lnt)+ 30[/tex]
(a)
First, we take its derivative
[tex]N'(t) = 20(2t-\frac{2}{t})[/tex]
Solve N'(t)=0
[tex]20(2t-\frac{2}{t})=0[/tex]
Simplifying
[tex]2t^2-2=0[/tex]
Solving for t
[tex]t=1\ ,t=-1[/tex]
Only t=1 belongs to the valid interval [tex]1\leqslant t\leqslant 15[/tex]
Taking the second derivative
[tex]N''(t) = 20(2+\frac{2}{t^2})[/tex]
Which is always positive, so t=1 is a minimum
(b)
[tex]N(1)=20(1^2-2ln1)+ 30[/tex]
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)
[tex]N(15)=20(15^2-2ln15)+ 30[/tex]
N(15)=4391.7 is a maximum