This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: So lets do the math to find out! ( for now on let A stand for area lost). A = 2500 + 50b A = 2500x + 50b A = 5000x. (A is positive). Since the second derivative of the area lost is 5000x, which is a positive number, we can now officially conclude that x = 50 is the global minimum value of b. b Now in order to find what the minimum number of firebreaks are for the four b values of our group, all we need to do is to plug the four b values into the equation x = 50 . b Here is what we got for each group members b value: Chales b value = .021km 50 for c= 101.42 (the minimal number of firebreaks for b Charles b value). Miyad b value = .01km, min. number of firebreaks = 70.21 (optimal # of firebreaks for all b values ). Tom b value = .016km, min. number of firebreaks = 88.6 Ray b value = .019km, min number of firebreaks = 96.5....
View
Full
Document
This note was uploaded on 04/28/2008 for the course CALC 101 taught by Professor Wiesner during the Spring '08 term at Ithaca College.
 Spring '08
 Wiesner
 Derivative

Click to edit the document details