Calculus Word ProblemsDate: 04/03/97 at 15:00:18 From: Nikita Bacchus Subject: Calculus Word Problems Hi, I really hope you can help me with my problems. I have tried to figure out these questions but I can't seem to get an answer. 1) a) For any numbers a1,a2,a3,a4 find the value of x that minimizes f(x)=(x-a1)^2 + (x-a2)^2+(x-a3)^2+(x-a4)^2. b) Using your answer to part (a) guess the value of x that minimizes f(x)=(x-a1)^2+(x-a2)^2++(x-a3)^2+...+(x-an)^2. 2) The cost of fuel to propel a boat through the water (in dollars per hour) is proportional to the cube of the speed. A certain ferry boat uses $100 worth of fuel per hour when cruising at 10 miles per hour. Apart from fuel, the cost of running this ferry (labor maintenance, and so on) is $675 per hour. At what speed should it travel so as to minimize the cost per mile traveled? Could you show me the steps in detail as to how to get the final answer, so that I could use it as a study tool? Thank you and I do hope to hear from you soon. Date: 04/03/97 at 17:10:29 From: Doctor Rob Subject: Re: Calculus Word Problems In part (1a), you need to differentiate f(x) with respect to x, and set the result equal to zero. The value of x that will make this differentiable function a minimum will be such that the derivative of the function will be zero there. To differentiate, you need to use the rules for the derivative of a sum and the derivative of a power, and you need to use the chain rule, since each square is a power of a function of x. d/dx(g(x)^n) = n*g(x)^(n-1)*g'(x) in general, and in your problem, g(x) = x-a1 in the first term, and g(x) = x-a2 in the second term, and so on. Then g'(x) = 1 foreach of these. Putting these all together, f'(x) = 2*(x-a1)^1*1 + 2*(x-a2)^1*1 + 2*(x-a3)^1*1 + 2*(x-a4)^1*1 = 2*[(x-a1)+(x-a2)+(x-a3)+(x-a4)] = 2*[4*x-(a1+a2+a3+a4)]. Setting this equal to zero, you can solve for x to get x = (a1+a2+a3+a4)/4. This means that the value of x that makes f(x) a minimum is just the mean (or average) of a1, a2, a3, and a4. For part (1b), why was the number 4 special in the solution of (a)? It wasn't! Question 2: "The cost of fuel to propel a boat through the water (in dollars per hour) is proportional to the cube of the speed." You want to write the total cost per mile as a function of the speed. Then you can differentiate with respect to the speed, set the derivative equal to zero (you want a maximum, and the derivative is zero at a maximum or minimum), and solve for the speed. Let F be the fuel cost in dollars per hour. Let v be the speed of the ferry in miles per hour. Let C be the total cost per mile. The first sentence says that F = k*v^3. "A certain ferry boat uses $100 worth of fuel per hour when cruising at 10 miles per hour." This sentence says that when v = 10, F = 100. Use these to find the constant k = 100/10^3 = 1/10 dollars*hour^2/miles^3. F = v^3/10. "Apart from fuel, the cost of running this ferry (labor maintenance, and so on) is $675 per hour." This sentence says the cost per hour of running the ferry is F + 675. To find the cost *per mile*, we have to divide by the speed in miles per hour. Thus C = (F+675)/v = v^2/10 + 675/v. "At what speed should it travel so as to minimize the cost per mile traveled?" Now we are ready to execute the plan outlined above: dC/dv = v/5 - 675/v^2 = 0 v^3 = 675*5 = 3375 v = 15 miles/hour F = 337.5 dollars/hour C = 22.5 + 45 = 67.5 dollars/mile. I hope this helps. If you need more detail, write again. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 04/03/97 at 18:07:43 From: Doctor Anthony Subject: Re: Calculus Word Problems >1a) For any numbers a1,a2,a3,a4 find the value of x that minimizes > > f(x)=(x-a1)^2 + (x-a2)^2+(x-a3)^2+(x-a4)^2. > > b) Using your answer to part (a) guess the value of x that minimizes > > f(x)=(x-a1)^2+(x-a2)^2++(x-a3)^2+...+(x-an)^2. dy/dx = 2(x-a1) + 2(x-a2) + 2(x-a3) + 2(x-a4) = 0 for minimum. 8x = 2(a1+a2+a3+a4) x = (1/4)(a1+a2+a3+a4), (b) the value of x to minimize f(x) if n terms are present is x = (1/n)(a1+a2+a3+........+an) >2)The cost of fuel to propel a boat through the water (in dollars per >hour) is proportional to the cube of the speed. A certain ferry boat >uses $100 worth of fuel per hour when cruising at 10 miles per hour. >Apart from fuel, the cost of running this ferry (labor maintenance, >and so on) is $675 per hour. At what speed should it travel so as to >minimize the cost per mile traveled? We must first find the constant of proportionality for cost of fuel against speed. Use the $100 per hour when travelling at 10 mph. 100 = k x 10^3 so k = 1/10. Let v be the speed of the boat. Then cost/hr = (1/10)v^3. Time = dist/speed, so for 1 mile, time = 1/v. The cost per mile in fuel = cost/hr x time = (1/10)v^3 x 1/v = (1/10)v^2. Labour cost per mile = cost/hr x time = 675 x 1/v = 675/v. Total cost per mile = (cost of fuel + cost of labour)/mile C = v^2/10 + 675/v For the minimum we find dC/dv = 2v/10 - 675/v^2 = 0 v/5 = 675/v^2 v^3 = 3375 v = 3375^(1/3) = 15 mph. So the boat should travel at 15 mph for minimum cost per mile. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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