Clear[a]

Plot[{2^x,2.1^x,(1/2)^x,(0.9)^x,x,x^2,x^4},{x,0,50}, PlotRange->{{0,30},{0,1}}]

.9^10

.9^100

Example 4.3 on page 279

Carbon-14 has a half life of 5770 years

Question 1:  If you start with 1 gram of carbon-14, how long does it take to have that 1 gram decay to 1/4 gram?

start with 1 gram, then 5770 years later, we have 1/2 gram.  Start with 1/2 a gram, 5770 years later, we have 1/4, so it will take 5770*2 years

Question 2:  if we start with 1 gram of carbon 14, how much is left after 1 year?

P(t) = N*(a^t)
Here N is the starting amount, and a is the growth/decay rate.  We know N is 1 gram. We do NOT know a.

Let’s assume, t is measured in years...  Then we know:

1/2 = 1*(a^5770)

5770*2

Solve[1/2==1*a^5770,a]

N[%]

NSolve[1/2 == a^(5770.0),a, Reals, WorkingPrecision->25]

P = 1*0.99987987772122755704594004835097049465`25.^t

P/.t->1

P/.t->5770

P/.t-> 100

Data = {{0,64},{1,45.44},{2,32.26},{3,22.91},{4,16.26},{5,11.5}};

Table[Data[[i+1]][[2]]/Data[[i]][[2]],{i,1,5}]

Model = A*Exp[B*t];

fit = FindFit[Data,Model,{A,B},t]

ModelFunction = Function[{t},Evaluate[Model/.fit]]

Plot[ModelFunction[t],{t,0,5}]

Problem 15 on 5.5

C(N) = fixed cost + cost per item *number of items
R(N) = Price(of item based on number of items)*number of items
Profit(N) = R(N) - C(N)

(a) C(N) = 2000 + 30*N

(b) use the data to find a LINEAR model for Price(N) i.e. for p(N)

WidgetData = {{200,41.00},{250,40.5},{300,40.00},{350,39.50}}

Table[(WidgetData[[i+1]][[2]]-WidgetData[[i]][[2]])/(WidgetData[[i+1]][[1]]-WidgetData[[i]][[1]]),{i,1,3}]

WidgetLine = LinearModelFit[WidgetData,{1,N},N]

so p(N) = 43-0.01*N

(c) R(N) = p(N)*N = (43-0.01*N)*N

R = (43-0.01*N)*N

Cost= 2000 + 30*N

Profit = R-Cost

Expand[%]

Plot[-2000+13 n-0.01(n^2),{n,0,1200}]

Solve[-2000+13 n-0.01(n^2)==0,n]