2.5 Inequalities

Example 2.10

The price of gold per ounce, G, can be expressed as a function t years after 2001 by:

G(t) = 255* (1.18^t)

G[t_]:=255*(1.18^t)

Part 1

To determine the price of gold in 2006, we note that 2006 is 5 years after 2001, which means we need to evaluate G(5), which is $583.38 per ounce as seen in the calculation below.

G[5]

Part 2

Plot[G[t],{t,0,9},AxesLabel->{"t years after 2001","$ per ounce"}, PlotRange->{{0,9},{0,1200}}]

Part 3

Here we want to know where $400 < G(t) < $700.

Part 4

Plot[{G[t],400,700},{t,0,7},AxesLabel->{"t years after 2001","$ per ounce"}]

NSolve[G[t]==400,t,Reals]

NSolve[G[t]==700,t,Reals]

The solution to the first equation G(t) = 400 is t=2.72 years after 2001, which would be the Fall of 2003, and the solution to the second equation G(t) = 700 is t=6.10 years after 2001, which would be the Spring of 2007.

Part 5

Plot[{G[t],900},{t,0,9},AxesLabel->{"t years after 2001","$ per ounce"}]

NSolve[G[t]==900,t,Reals]

So after t=7.62 years after 2001, which is the middle of 2008, the price of gold will remain above $900.

Example 2.11

A fish farm sells fingerlings and the income and expense of raising these little fish can be modeled by the following equations:

IN(n) = 5000(0.3n-0.02(n^2))
EX(n) = 1000+250n

Here the number of fish n is measured in hundreds, and the income and expense (IN and EX) are measured in $ per month.

IN[n_]:= 5000*(0.3n-0.02(n^2))

EX[n_]:=1000+250n

Part 1

TableForm[Table[{n,IN[n]},{n,0,20}],TableHeadings->{None,{"n hundred fish","Income $/month"}}]

Part 2

Plot[IN[n],{n,0,15}, AxesLabel->{"n hundred fish","Income $/month"}]

FuncPlots=Plot[{IN[n],EX[n]},{n,0,15}, AxesLabel->{"n hundred fish","Income $/month"}, PlotLegends->{"Income","Expense"}]

ShadePlot = RegionPlot[IN[n]>y && y > EX[n],{n,0,12},{y,1000,6000}, PlotPoints->60, PlotStyle->LightOrange,BoundaryStyle->None, Axes->True];

Show[ShadePlot,FuncPlots]

To be profitable, we need the income > expense, which, graphically, is where the blue line lies above the yellow line.  The shaded region in the figure above shows this area.  We can use the NSolve command to find these intersection points.

NSolve[IN[n]==EX[n],n,Reals]

The only two solutions to the NSolve command are the two where the curves intersect above.  The profitable region is 0.86 < n < 11.64 hundreds of fish.

Exercise 3

The formula for the amount of cyanide in the bloodstream after smoking a cigarette us given by
C(t) = 0.1 + 0.3 (t^0.6)(e^(-0.17t))
Here t is being measured in minutes, and the concentration of cyanide C is measured in ng/dl.

CY[t_]:=0.1+0.3(t^0.6)(E^(-0.17t))

Part a.

Plot[{CY[t],0.3},{t,0,60},AxesLabel->{"time (min)", "Cyanide (ng/dl)"}]

Part b.

We see from the graph above that the concentration goes above the 0.3 ng/dl concentration somewhere right after both t=0 and t=10.

NSolve[CY[t]==0.3,t,Reals]

From the NSolve command, the range where levels are above the target range is 0.60 < t < 10.78 minutes.

Exercise 7

Wave height h, in feet, can be estimated by wind speed w in miles per hour (mph) using the formula
h(w) = 0.02(w^2)

h[w_]:=0.02(w^2)

Part a.

Plot[h[w],{w,0,25},AxesLabel->{"wind speed (mph)", "wave height (ft)"}]

Part b.

Plot[{h[w],4},{w,0,15},AxesLabel->{"wind speed (mph)", "wave height (ft)"}]

NSolve[h[w]==4,w,Reals]

The range of speeds will be 0 < w < 14.14 mph since wind speeds cannot be negative.