# Max/Min Online Webwork Problem

• Apr 10th 2007, 12:07 PM
qbkr21
Max/Min Online Webwork Problem
Here is the question:

http://item.slide.com/r/1/27/i/gWl26...JpL_l1TxLHaER/

Thus I went about trying to solve this problem in this manner. Note that the Minimum value is correct but the system will not accept 10 as an answer for the maximum value.

http://item.slide.com/r/1/101/i/giky..._QLLTPBa18NOb/
• Apr 10th 2007, 12:17 PM
Jhevon
Quote:

Originally Posted by qbkr21
Here is the question:

http://item.slide.com/r/1/27/i/gWl26...JpL_l1TxLHaER/

Thus I went about trying to solve this problem in this manner. Note that the Minimum value is correct but the system will not accept 10 as an answer for the maximum value.

http://item.slide.com/r/1/101/i/giky..._QLLTPBa18NOb/

remember that absolute max is different from local max. the derivative gives you the local max. the absolute max is the highest point in the interval, IT DOES NOT HAVE TO BE A CRITICAL POINT.

check the end points, we get:

(15, 28585) and
(-6, -2222)

so the absolute max is 28585
• Apr 10th 2007, 12:19 PM
qbkr21
Quote:

Originally Posted by Jhevon
remember that absolute max is different from local max. the derivative gives you the local max. the absolute max is the highest point in the interval, IT DOES NOT HAVE TO BE A CRITICAL POINT.

check the end points, we get:

(15, 28585) and
(-6, -2222)

so the absolute max is 28585

Bingo You were right, but what did you do to maximize each coordinate? Did you stick the intervals that x was between back into f(x)?
• Apr 10th 2007, 12:21 PM
Jhevon
Quote:

Originally Posted by qbkr21
Bingo You were right, but what did you do to maximize each coordinate? Did you stick the intervals that x was between back into f(x)?

yes, i found f(-6) and f(15). if one of those is lower than the y-value of all critical poitns, then it is the absolute min, if one is higher than all the y-values of the critical points, it is the absolute max.

so always remember to check the endpoints for absolute max and mins