1)A forest in shape of 50 km by km squared has firebreaks in rectangular strips
ID: 2690463 • Letter: 1
Question
1)A forest in shape of 50 km by km squared has firebreaks in rectangular strips 50km by .01 km. The trees between two firebreaks are called a stand of trees. All firebreaks in this forest are parallel to each other and to one edge of the forest, with the first and last firebreaks at the edges of the forest. The firebreaks are evenly spaced throughout the forest. The total area lost in the case of a fire is the area of the stand of trees in which the fire started plus the area of all the firebreaks. a) Find the number of firebreaks that minimizes the total area lost to the forest in the case of a fire. b) If a firebreak is 50 km by b km, find the optimal number of firebreaks as a function of b. If the width, b, of a firebreak is quadrupled, how does the optimal number of firebreaks change? 2)Now suppose firebreaks are arranged in two equally spaced sets of parallel lines, one set vertical and the other set horizontal. The forest is a 50 km by 50 km square, and each firebreak is a rectangular strip 50 km by 0.01 km. Find the number of firebreaks in each direction that minimizes the total area lost to the forest in the case of a fire. ANY HELPExplanation / Answer
Interesting problem... unless you forgot to type something ... part 1 and 2 makes more " Calculus" sense than part b)... Let the area of a firebreaks be (.01)(50 = .5 sq km and the area of the entire forest = 50 (50 ) = 2500 sq km now make a series of sketches : Part 1) start with two firebreaks on each edge : the amount thats lost is 2(.5) + (1(2500) [ it all burns ] now add a strip down the middle : amount lost = 3(.5 ) +(2500) /2 [ only one half burns ] and another firebreaks : A = 4(.5 ) + 2500/3 and so on ... If we do real calculus, and consider the firebreaks to be an almost negligible part of the forest, unless we get up to 50 firebreaks, then the general amount lost is : A = x( .5 ) + 2500 / ( x-1) where x is the number of firebreaks and A , I think, has to be greater than 0 ...if there is a fire .. so A = x(.5) + 2500 /(x-1 )Related Questions
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