Ask Your Question

Help to calculate max with CAS

asked 2016-06-16 23:39:20 -0500

this post is marked as community wiki

This post is a wiki. Anyone with karma >750 is welcome to improve it.


I need your help please. I am in the last step (step 3) of finishing this paper for my graduate degree, and I do not know what is a CAS or how to use one. Here is my function. I have already determined the maximum population of bacteria using a graphing calculator and saved that image, but I need to repeat the process on a CAS and show an image of it. SAGE has been referred to as the best, but I have no idea where to begin.

The number N of bacteria in a culture at time d, where d is measured in days, is modelled with the following function:

N = (12250d^2 - 2419d + 433768) / (5d^2 -37d+99)

A. Use technology to analyze the modeling function by doingthe following: 1. Use a calculator to determine the number of bacteria whend = 10 N = 7,137.8952bacteria where d = 10

  1. Use a graphing utility to graph and estimate the maximumnumber of bacteria. a. Provide an image of the graphproduced by the graphing utility N = 20,116.296maximum bacteria

  2. Determine the exact value for the maximum number ofbacteria, using a computer algebra system. a. Provide an image of the output fromthe computer algebra system.

edit retag flag offensive close merge delete

1 answer

Sort by ยป oldest newest most voted

answered 2016-06-19 14:21:09 -0500

eric_g gravatar image

Here is a possible solution:

sage: var('d')  # declare d as a symbolic variable
sage: N(d) = (12250*d^2 - 2419*d + 433768) / (5*d^2 -37*d+99)
sage: plot(N(d), (d,0,10), axes_labels=['d', 'N'])  # the graph of N as a function of d
Launched png viewer for Graphics object consisting of 1 graphics primitive
sage: der = diff(N(d), d).simplify_full() ; der  # the derivative N'(d)
-5*(88231*d^2 + 382436*d - 3161987)/(25*d^4 - 370*d^3 + 2359*d^2 - 7326*d + 9801)
sage: sol = solve(der==0, d, solution_dict=True)  # searching for the zeros of the derivative
sage: sol
[{d: -1/88231*sqrt(315549598521) - 191218/88231},
 {d: 1/88231*sqrt(315549598521) - 191218/88231}]
sage: d_max = sol[1][d] ; d_max  # among the above two roots, we select the positive one
1/88231*sqrt(315549598521) - 191218/88231
sage: n(d_max)  # a numerical approximation of the number of days at which max(N) is achieved
sage: N_max = N(d_max).simplify_full() ; N_max  # the maximum number of bacteria
3*(1632757263*sqrt(315549598521) - 2576988387921500)/(5176727*sqrt(315549598521) - 3155495985210)
sage: n(N_max)  # a numerical approximation of this number
edit flag offensive delete link more

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools


Asked: 2016-06-16 23:39:20 -0500

Seen: 215 times

Last updated: Jun 19 '16