@Kitonum Okey, that worked, now the graph looks nice and understandable..thank you for your help!

@Kitonum Oh dear, you're right..I tried to write my Maple command on a seperate worksheet and it turned out fine. Hermmm..but thank you anyways!

@Kitonum Yes I managed to get a similar graph to yours with this Maple command

f := x -> (5*x)/(2+x);

g := x -> (5*x)/(3+x);

plots:-display(
  plot( [f(x),g(x)], x=0..30, color=[red,blue], legend = ["k=2", "k=3"]));

But I was wondering whether I can make it so that the point k=2 and k=3 is marked on the graph itself? Hermm, like dashed lines from the graph going down to where x=2?
 

@Carl Love But when I tried to find the upper and lower bounds of the graph manually, that is N'(t)=0 and differentitating N(t) , I got t=1.2793 and 4.4209 which is the supposed answer.

N'(t)= d/dt (10+ 2e^-0.3t sin t)=0

      = 0 + 2e^-0.3t cos t -0.6e^-0.3t sin t=0

                                   2 cos t= 0.6 sin t

                                      tan t= 2/0.6

                                            t= tan^-1 20/6

                           Thus, t= 4.4209 and t=1.2793

This is making me so confused. Manually, the answer is correct according to the answer sheet given by my lecturer but Maple gives a very different graph.

@Carl Love 

For b)

R=aP^2/3

For P>0, R/P= aP^2/3/P, simplify using law of exponents, then R/P= aP^-1/3

Since function R/P= aP^-1/3 is a power function, it is continuous whenever it is defined, since it is defined for P>0, then it is continuous for P>0.  Thus, R/P as a function of P is continuous for P>0.

lim_P→0+   R/P = lim _P→0+aP^-1/3 =lim a/P^1/3=∞              since a>0 and P^1/3 →0+ as    P→0+

Same goes for lim_P→∞ R/P=0
   

@Carl Love Not homework, just review questions for my course. I'll be having an exam this Friday and am currently doing exercises.  The thing is, manually I am able to find the answers. It's  just that I've never actually learned Maple before. If you've noticed, all the questions I've posted were all related to Maple.

For a)

R= aP^b

Since the quote specifies the power is 2/3, then b must be 2/3. Then R= aP^2/3

Taking log of both sides, then log R= log a + b log P which is equivalent to log R= b log P + log a

Theoretically, the graph would be a straight line with slope b and intercept log a. By graphing log R versus log P, the slope can be shown to be 2/3 ( which indicates b approximately 2/3). 

But, I just don't know how to use Maple.