Anyone a math genius? - DPChallenge Forums

04/05/2009 05:15:15 PM · #1

If anyone can answer these questions that'll be cool.

1) A manufacturer is predicing profit, P, in thousands of dollars, on the sale of x tonnes of fertilizer according to the equation P(x) = 600x - 15000/x+100
- Describe the predicted profit as sales increase
- Compare the rates of change of the profit at sales of 100 t and 500 t of fertilizer

2) The electric power, P, in watts, delivered by a certain battery is given by the function P = 100R/(2+R)^2, where R is the resistance in ohms.
- Describe the power output as the resistance increases from 0 ohms to 20 ohms
- Show that the rate of change is 0 at R = 2. What does this indicate about the power?

Message edited by author 2009-04-05 17:22:05.

Canon RF 24-70mm f/2.8 L IS

04/05/2009 05:26:19 PM · #2

is that your math homework due tomorrow? lol
you don't have to be a genious to solve these though...do you have a graphic calculator you can visualize this with?
i might give it a try just to prove to myself i can still do this kinda stuff...;)

kirbic

Canon EOS R5

04/05/2009 05:28:05 PM · #3

Do you have Microsoft Excel? You've already typed pretty much what you need to type to get the answers.
Example: for the first question, enter the value for x (the amount sold) in cell A1. In cell A2, enter this formula:
=600*A1-15000/A1+100
Now change your input (x) as desired and the profit (P) will display in cell A2.

You can do the same with the other equation. Hint on the rate of change question: rate of change is zero where the slope of the graph of power vs. resistance is zero.

04/05/2009 05:40:12 PM · #4

ok i dunno if that helps but the quadrativ equation will look like this:
0=600xÂ²+100x-15000
x(1)=4,88

that means that when the sales exceed 4,88t of fertilizer the guy will make profit.

04/05/2009 05:51:41 PM · #5

thanks guys. Yeah well basically this is kind of an assignment and it's due tomorrow and i've been trying to figure it out all day. I'm pretty bad at math, but what makes it worse is that we have to figure these out on our own. The whole entire chapter had to be self-taught, so i know as much as someone who has never done math before.

I'm still kinda unsure what rate of change is :S

04/05/2009 06:16:48 PM · #6

P(x=100)=59500
P(x=500)=300070

x(500)/x(100)=5:1
P(x=500)/P(x=100)=5,04:1

that means the increase of profit is almost proportional with the increase of sales.

eta: please take everything i'm saying with a good amount of salt, you know it's been 3 years since my last math lessons...;)

btw the second problem is a lil tougher, cus you need a special formula to kinda convert that equation to a 0=aXÂ²+bX+c equation. or something like that... :D

Message edited by author 2009-04-05 18:22:40.

04/05/2009 06:45:00 PM · #7

oh okay thanks :)

04/05/2009 07:37:59 PM · #8

Originally posted by cujee:

Rule 1: Derivative of a function gives its rate of change.

So lets say P(x) = 600x - 15000/x+100 = Ax - B/x + C

or
x.P = Ax^2 -B + Cx

take derivative:
P.dx/dx + x dPdx = 2Ax + C

or
P + xdP/dx = 2Ax + C

or
(Ax - B/x + C) + xdP/dx = 2Ax + C

or
xdP/dx = Ax + B/x

This is your rate of change of profit:
dP/dx = A + B/(x^2)

Put the value of x and you get comparison.

Similar things you could do for other functions.

Message edited by author 2009-04-05 19:38:55.

04/05/2009 07:50:08 PM · #9

ok i sort of figured it out...

so the power output (P) seems way high when the resistence (R) is near 0, then decreases to about 120W when R=2ohms (that's the turning point, meaning that the rate of change of the resistence is 0!) and from there increases and continues so. the power output reaches 530W when the resistence is 20ohms.

here's the converted equation and the visualized graph:

that was kinda fun actually! :D

Message edited by author 2009-04-05 20:11:29.

04/05/2009 07:52:10 PM · #10

Since this is your homework, I probably shouldn't do this, but you got my interest piqued.

If I did this right (and I could be wrong, probably am, and if I wanted to teach you a lesson about possible cheating, I probably did so on purpose) I came up with the following:

It's up to you to decide if I'm right or wrong. *evil grin*

EDIT: I did screw up!!!! Don't look at the second graph or you're going to be screwed up too!!!

Message edited by author 2009-04-05 19:55:49.

04/05/2009 08:01:32 PM · #11

Let's try this one instead (again, I could be wrong and probably am.)...

04/05/2009 08:04:40 PM · #12

ah crap i think i screwed up my graph as well...gimme a second to figure out what's wrong.

eta: i can confirm with nathan's graph now.
sure glad i still have my good old 32kb graphic calculator...:D

Message edited by author 2009-04-05 20:09:45.

Nikon AF Zoom-Nikkor 35-135mm f/3.5-4.5

04/05/2009 08:09:49 PM · #13

I think we didn't pay enough attention in class all those years ago. *grin*

C_Steve_G

Apple iPhone 12

04/05/2009 08:11:24 PM · #14

NathanW, your second graph is correct. I'm wondering what this formula is actually trying to demonstrate other than setting up a math problem...
It's similar to the charactaristic of a battery's ability to provide power depending on its internal resistance, which has a relation to its state of charge.

Message edited by author 2009-04-05 20:18:31.

04/05/2009 08:21:01 PM · #15

don't torture yourself guys:

Go to this page:

//integrals.wolfram.com/index.jsp

And enter
A + B/(x^2)

(the rate that i derived)

And integrate.

You get back your original function (without the constant off course).

So what i derived is correct answer.

:-D

Canon EF-S 18-200mm f/3.5-5.6 IS

04/05/2009 08:23:02 PM · #16

Originally posted by C_Steve_G:

NathanW, your second graph is correct.

I've actually gotten pretty good at using excel to do my math for me. The problem with the first graph was that I didn't enter the equation correctly.

I don't have a clue how I'd use either of these for much of anything I do on regular/irregular basis. But then I've also found that a lot of the stuff I'm learning in my MBA doesn't pertain to most of the businesses in the area I live. *shrug* Good thing I like to learn.

CEJ

Canon EOS-40D

04/05/2009 08:38:19 PM · #17

Don't feel too bad...NOTHING I learned acquiring my MBA has anything to do with my current practical application of the knowledge.

04/05/2009 08:53:04 PM · #18

Originally posted by zxaar:

Originally posted by cujee:
If anyone can answer these questions that'll be cool.

1) A manufacturer is predicing profit, P, in thousands of dollars, on the sale of x tonnes of fertilizer according to the equation P(x) = 600x - 15000/x+100
- Describe the predicted profit as sales increase
- Compare the rates of change of the profit at sales of 100 t and 500 t of fertilizer

2) The electric power, P, in watts, delivered by a certain battery is given by the function P = 100R/(2+R)^2, where R is the resistance in ohms.
- Describe the power output as the resistance increases from 0 ohms to 20 ohms
- Show that the rate of change is 0 at R = 2. What does this indicate about the power?

Rule 1: Derivative of a function gives its rate of change.

So lets say P(x) = 600x - 15000/x+100 = Ax - B/x + C

or
x.P = Ax^2 -B + Cx

take derivative:
P.dx/dx + x dPdx = 2Ax + C

or
P + xdP/dx = 2Ax + C

or
(Ax - B/x + C) + xdP/dx = 2Ax + C

or
xdP/dx = Ax + B/x

This is your rate of change of profit:
dP/dx = A + B/(x^2)

Put the value of x and you get comparison.

Similar things you could do for other functions.

What's the "d" suppose to represent?

Canon EF-S 18-135mm f/3.5-5.6 IS STM

04/05/2009 09:01:11 PM · #19

at the back of the book it says the rate of change for the profit and fertilizer one:

the rate of change of the profit at 100 t is 1.875 and approximately .0208 at 500 t, so the rate of change is decreasing.

pedrobop

Canon PowerShot SX100 IS

04/05/2009 09:26:00 PM · #20

This thread was really great to remember me how much i hate math!

04/05/2009 09:30:39 PM · #21

Originally posted by cujee:

Originally posted by zxaar:

Originally posted by cujee:
If anyone can answer these questions that'll be cool.

1) A manufacturer is predicing profit, P, in thousands of dollars, on the sale of x tonnes of fertilizer according to the equation P(x) = 600x - 15000/x+100
- Describe the predicted profit as sales increase
- Compare the rates of change of the profit at sales of 100 t and 500 t of fertilizer

2) The electric power, P, in watts, delivered by a certain battery is given by the function P = 100R/(2+R)^2, where R is the resistance in ohms.
- Describe the power output as the resistance increases from 0 ohms to 20 ohms
- Show that the rate of change is 0 at R = 2. What does this indicate about the power?

Rule 1: Derivative of a function gives its rate of change.

So lets say P(x) = 600x - 15000/x+100 = Ax - B/x + C

or
x.P = Ax^2 -B + Cx

take derivative:
P.dx/dx + x dPdx = 2Ax + C

or
P + xdP/dx = 2Ax + C

or
(Ax - B/x + C) + xdP/dx = 2Ax + C

or
xdP/dx = Ax + B/x

This is your rate of change of profit:
dP/dx = A + B/(x^2)

Put the value of x and you get comparison.

Similar things you could do for other functions.

What's the "d" suppose to represent?

I am not sure what standard you are in.

I assume in india 11st of student should understand that d stands for derivative.

Probably i would have written it in more simpler way. If there is any.

04/05/2009 09:45:27 PM · #22

lol, clearly the standard is much lower. No but i'm awful at math, and i have to self-teach this unit myself so you have to explain it in the most simplistic way possible.

04/05/2009 10:00:22 PM · #23

Originally posted by cujee:

lol, clearly the standard is much lower. No but i'm awful at math, and i have to self-teach this unit myself so you have to explain it in the most simplistic way possible.

Derivative is nothing but rate of change of that function.
Imagine that you write a function of y=f(x).
Then df(x)/dx = rate of change of that function at any given x.

If you want to avoid derivative completely then only go with the basic defination that is 'rate of change'.

Now you go to ask yourself what is rate of change and how can i calculate this.
A very simple way to estimate rate of change is :
rate of change = (f(x2) - f(x1))/ (x2 - x1).

This gives you an estimation of rate of change around (x1 + x2) / 2 . (that is mid point of x1 and x2).

In excel if you want to see this in action.

Plot
x , f(x).

Now in next column create rate of change function by formula i gave you.
For this for ith row in excel use function values of i+1, and i-1 row. (similarly x of i+1 and i-1 row).

Plot this function with respect to x and you get your rate of function.

If question is about behaviour of rate of function, just show this plot to your teacher and tell him this is how it shall behave. (you are done here).

If rate of function is asked at some x1, x2, x3 etc.

Just use that chart to predict it and answer that. (or you can interpolate).

ps: in doubt always ask.