A smokestack deposits soot on the ground with a concentration inversely proporti

Question

A smokestack deposits soot on the ground with a concentration inversely proportional to the square of the distance from the stack. With two smokestacks d miles apart, the concentration of the combined deposits on the line joining them, at a distance x from one stack, is given by S = (c/x^2)(k/(d-x)^2) where c and k are positive constants which depend on the quantity of smoke each stack is emitting. If k = 6 c, find the point on the line joining the stacks where the concentration of the deposit is a minimum

Explanation / Answer

A smokestack deposits soot on the ground with concentration inversely proportional to the square of the distance from the stack. With two smokestacks 20 miles apart, the concentration of the combined deposits on the line joining them, at a distance x from one stack, is given by the following equation where k1 and k2 are positive constants which depend on the quantity of smoke each stack is emitting. S= (k1/x^2) + (k2/(20-x)^2) If k1 = 8k2, find the point on the line joining the stacks where the concentration of the deposit is a minimum. so first rewrite the equation for S replacing k1 w/ 8k2. as w/ most minimization problems, all you have to do is take the derivative (w/ respect to x) and set it equal to zero. you should get something like 0 = k2*f(x) where f(x) is some messy function of x which i don't really want to solve for you. but you don't know what k2 is right? oh well, just divide it out b/c you can assume that it's not zero, b/c otherwise 1) there wouldn't be any emission from the stacks and 2) all x's would be minima :-P so you should get 0 = f(x)... and f(x) should be nice enough that you can solve for x. note if you can get it in the form of something like g(x)/h(x)... then all you have to do is set g(x) = 0 b/c you're only looking to find the minimum for values of x b/t 0 and 20 A smokestack deposits soot on the ground with a concentration inversely proportional to the square of the distance from the stack. With two smokestacks 20 miles apart, he concentration of the combined deposits on the line joining them, at a distance [x] from one stack is given by, S = (k1/x^2) + (k2/[20-x]^2) where k1 and k2 are positive constants which depend on the quantity of smoke each stack is emitting. If k1 = 7k2, find the point on the line joining the stacks where the concentration of the deposit is a MINIMUM. ake the derivative of S and set it equal to 0. Use the fact that k1 = 7*k2, and you'll find that the answer does NOT depend on the actual value of k1, because k1 just factors out. The minimum obviously doesn't occur at an endpoint, because S goes to infinity at the endpoints. So where else can the minimum be A smokestack deposits soot on the ground with a concentration inversely proportional to the square of the distance from the stack. With two smokestacks 20 miles apart, the concentration of the combined deposits on the line joining them, at a distance x from the first stack (closest to New York City), is given by S=84/x2 + 12/(20-x)2 ake derivative and set to zero. -168/x^3+24/(20-x)^3=0 x=13.134 2. xy=2300 Cost = 12x+21x+21*2*y =12x+21x+21*2*(2300/x) =33x+96600/x Minimize, so take derivative... 33-96600/x^2=0 33x^2=96600 x=54.10 feet So y=42.51 feet

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