A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of water is approximated byN(t) = 20(t12−ln(t12))+ 30for 1≤t≤15.a. When will the number of bacteria be a minimum?b. What is the minimum number of bacteria during this time?c. When will the number of bacteria be a maximum?d. What is the maximum number of bacteria during this time?

Answer:N(1)=50 is a minimum N(15)=4391.7 is a maximumStep-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