# Homework Help: Evaluating the instantaneous rate of change using h = 0.1

1. Jul 19, 2013

### needingtoknow

1. The problem statement, all variables and given/known data

Write a simplified algebraic expression, in terms of a and h, to estimate the instantaneous rate of change of the profit for each numeber of athletic shirts sold. Use h = 0.1.

3. The attempt at a solution

{ -0.25(1000+h)^2 + 4(1000+h) + 12 - [ -0.25(1000)^2 + 4(1000) + 12] } / h
{ -0.25(1000000+2000h + h^2) + 4000 + 4h + 12 - [-245 988] } / h
All the constants cancel out to zero which leaves me with
(-0.25h^2 - 496h)/h = -0.25h-496

The when I plug in h = 0.1. I get 496.025 when the answer in the back states 3.475. What am I doing wrong?

2. Jul 19, 2013

### Staff: Mentor

Where does the 1000 come from? Presumably, it's from the number of shirts sold.

It would be helpful if you gave us all of the given information, including the profit function. It's possible that the variable in the profit function represents the number of shirts sold, in thousands.

Last edited: Jul 19, 2013
3. Jul 19, 2013

### verty

If the profit for x shirts is given by $- \frac{1}{4}x^2 + 4x + 12$ then I don't see anything wrong. The rate of change could never be ~3.

Hmm, Mark44 makes a very good point, a larger secant could indeed give a lower rate of change.

4. Jul 19, 2013

### HallsofIvy

needingtoknow, the point is that you don't tell us what function you have or what the number of shirts is at which you are trying to find the derivative. Mark44 guessed that the number of shirts was 1000 and verty guessed that the function was -(1/4)x^2+ 4x+ 12. Please tell people things like that! Don't make them guess.

5. Jul 19, 2013

### needingtoknow

Oh sorry I missed to include that. The profit function is P(n) = -0.25n^2 + 4n + 12 where n is the number of athletic shirts sold, in thousands and P ios the profit in thousands of dollars.

Last edited: Jul 19, 2013
6. Jul 19, 2013

### needingtoknow

and yes 1000 is the number of athletic shirts sold

7. Jul 19, 2013

### Staff: Mentor

Which means that, as I suspected, n should be 1 in your work, not 1000.

You need to work with this: { -0.25(1+h)^2 + 4(1+h) + 12 - [ -0.25(1)^2 + 4(1) + 12] } / h

8. Jul 19, 2013

### needingtoknow

Oh yes that fixed it, thank you very much! I had a similar problem with an earlier question perhaps it is for the same reason. Thank you!

9. Jul 20, 2013

### HallsofIvy

By the way, I notice that you titled this "Evaluating the instantaneous rate of change using h= 0.1" but the problem actually asks you to estimate it. Those are very different things. You cannot, in general, "evaluate" a rate of change using a specific value of h- you have to take the limit as h goes to 0.

Last edited by a moderator: Jul 20, 2013