course Mth 173 Calculus I12-05-2007
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17:55:47 Query 4.5.8 (problem 6 p 275) 3000 ft^2 rect region costing $25/ft on 3 sides and $10/ft on fourth; min cost. What is the minimum total cost?
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RESPONSE --> 3000= L*W 3000= D * d 3000/ D = d L is a compostion of 2 sides of $10 and $25 which gives L= $35 / ft. W= $50 / ft Since W costs 70% more than L we the maximum effciency of the function will be when W=.7L C= D(35) + d(50) C= D(35) + (3000/ D) (50) C= D(35) + (150000 / D) This gives us the cost for the area, if we set the derivative to 0 then we will get a critical point in the graph which should be a minima in the cost graph. C'= 35 - (150000 / D^2) 0= 35 - (150000/D^2) 35 = 150000/D^2 35 * D^2 = 150000 D^2 = 4285.7 D= 65.465 This means L= 65.465 and W= 45.826 which confirms the hypothesis of the minima being at W=.7L and sets the price at $4,582.58
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17:58:40 If x stands for the length of the fourth side of the region and y for the other side then we have area = x * y = 3000, so that y = 3000 / x. The cost of fencing the two sides of length x are $10 / ft and $25 / ft, for a total of 10 x + 25 x = 35 x. The cost of fencing the two sides of length y is $25 / ft for a total of 25 ( 2y) = 50 y. The total cost is therefore 35 x + 50 y. Using the fact that y = 3000 / x we find that total cost = C(x) = 35 x + 50 ( 3000 / x) = 35 x + 150000 / x. The minimum cost is found by minimizing this expression. We first find the derivative C ' (x) = 35 - 150000 / x^2. Setting this derivative equal to zero we have 35 - 150000 / x^2 = 0. Multiplying both sides by x^2 we obtain 35 x^2 - 150000 = 0 so that x^2 = 150000 / 35 and x = sqrt(150000 / 35) = 65, approx.. We see that this is a minimum, since the second derivative of C(x) is C ''(x) = 450000 / x^3, which is positive for all positive x. The other side is y = 3000 / x = 3000 / 65 = 46, approx..
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RESPONSE --> Where you used the second derivative I used the W=.7L which served the same purpose. It made sure that the critical point given by the derivative was in fact the minima of the cost function.
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18:12:02 Query 4.5.21 (was problem 18 p 276 ) towns at (0,-1) and (4,-4), river on x axis; find min dist of pipe to supply both. What is the minimum total length of pipe?
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RESPONSE --> Town {(0,-1), (4, -4)} River = x axis To find the minimum distance of pipe to supply both towns. There are three possibilites, running straight to the river from the towns (5 units) and going to from the river to (0,-1) then to (4, -4) (6 units 5 of which are used between the two towns) Therefore it is essentially pointless the persue the thrid possibility when a verticle pipe is run from the ""stretcher"" between the towns to the river, since just the ""stretcher"" pipe is as much as each town just having their own verticle pipe. So it is reasonable to say that the minimum amount of pipe is 5 units with each town having it's own pipe straight to the river.
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18:25:16 If the pipe runs from a point x units along the river from the closest point to the first town, then the pipe running to the first town will have length sqrt(1^2 + x^2) and the pipe to the second town will have length sqrt(4-x)^2 + 4^2), so total pipe length will be L(x) = sqrt(1^2 + x^2) + sqrt(4^2 + (4-x)^2). The derivative of this expression is L ' (x) = x / sqrt(1 + x^2) - ( 4 - x) / sqrt(16 + (4-x)^2). Setting this expression equal to 0 and multiplying by the common denominator sqrt(1+x^2) * sqrt( 16 + (4-x)^2 ) we obtain the equation x sqrt(16 + (4-x)^2 ) + ( x - 4) sqrt(1 + x^2) = 0. Placing the two expressions on opposite sides of the = sign we obtain sqrt(16 + (4-x)^2 ) = -( x - 4) sqrt(1 + x^2). Squaring both sides we obtain x^2 ( 16 + (4-x)^2 ) = (x-4)^2 (1 + x^2). Expanding we have x^2 ( 32 - 8 x + x^2 ) = ( x^2 - 8 x + 16 ) (1 + x^2), and further expanding we have x^4 - 8·x^3 + 32·x^2 = x^4 - 8·x^3 + 17·x^2 - 8·x + 16. Subtracting x^4 - 8 x^3 from both sides gives us 32 x^2 = 17 x^2 - 8 x + 16, which we simplify to the basic form of a quadratic function 15 x^2 + 8 x - 16 = 0. This function has two solutions, x = .8 and x = -4/3. We reject the second solution and accept the first, concluding that L ' (x) = 0 when x = .8. A second derivative test or a first derivative test tells us that L(x) indeed has a minimum at x = .8. The total length of pipe is therefore L(.8) = 6.4 miles, approx..
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RESPONSE --> The problems didn't match up in my book so I may have missed something such as the scale, and something saying that the pipe must only reach the river at one point. To find the distance between the towns I used the distance formula 'sqrt((Y-y)^2 + (X-x)^2)
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18:39:05 Query 4.6.16. (was problem 12 p 280 ) family a cosh(x/a), a>0. Describe your graphs.
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RESPONSE --> cosh = (e^(x/a) + e^(-x/a)) / 2 when a>0 As +- x becomes larger the functions near 0 because they are asymptotic to the x axis. As a becomes larger it makes the slope of the function smaller, making the functions wider in the hyperbola.
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18:40:35 ** a cosh(x/a) = a ( e^(x/a) + e^(-x/a) ) / 2. The exponential functions are reflections of one another about the y axis. Both are continuous at x=0, both equal 1 at x = 0. So at x = 0 the function y = a cosh(x/a) takes value a ( 1 + 1) / 2 = a. The function is increasing for positive x, and since for large x the expression e^(-x/a) approaches 0 (as a result of a being positive), a cosh (x/a) becomes very close to the exponential function a e^(x/a) as x increases. For increasing a, for any given positive x the slope sinh(x/a) becomes less since x/a becomes less and sinh is an increasing function for positive x. For negative x the graph is the reflection through the y axis of the positive-x graph. The function is everywhere continuous and differentiable. **
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RESPONSE --> okay
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18:46:58 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> In this we're using the calculus that we have learned to figure out something that you know can be figured out but it just takes some reconfiguring of the variables to make them able to be put into a function and figured out using optimization.
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