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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....
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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