course Mth 173 I am terribly sorry, but I've had a difficult time with the assignments this week. The testing center at WCC closes just before lunch on Fridays, and I am afraid I don't have enough time to finish the last assignment and take the test before they close tomorrow morning. I'll do my best to take it first thing Monday morning. Again, I'm sorry about the delay. Thanks for your patience with me this semester! Z?I???????????assignment #028s?r??????????????Calculus I
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22:17:45 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 --> 4th edition Problem 4.5.19, not assigned. First off, I need to minimize costs for a 3000 ft^2 region. I can get a simple relationship by the area of the box: xy = 3000 The next relationship isn't quite as simple, as it relates the cost. I have chosen to multiply the two x sides and one y side by the cost of $25, leaving the other y side for the cost of $10. That makes the relationship: Cost = 50x + 25y + 10y Now I can solve for y in the simple equation: y = 3000 / x Substitute: C = 50x + 25(3000 / x) + 10(3000 / x) Simplify: C = 50x + 105000 / x Now I need to find the derivative: C' = 50 - 105000 / x^2 Solve for all critical points: 0 = 50 - 105000 / x^2 50 = 105000 / x^2 x^2 = 105000 / 50 x^2 = 2100 x = 'sqrt(2100) x = 45.8 (-45.8 feet per side would be meaningless) Now that I have x, all I need to do is solve for y: y = 3000 / 45.8 = 65.5 Now plug into the original Cost equation to find cost: Cost = 50(45.8) + 25(65.5) + 10(65.5) Minimum cost = $4582.50
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22:18:19 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 --> Ok.
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22:29:43 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 --> Problem 4.5.32. I need to minimize pipe length, so I need to find a formula for pipe length and take the derivative. So the first pipe length is: P1 = 'sqrt(1^2 + x^2) The second pipe length is: P2 = 'sqrt(4^2 + (4-x)^2) Total length is P1 + P2: T = 'sqrt(1^2 + x^2) + 'sqrt(4^2 + (4-x)^2) I found the derivative to be: T' = x / 'sqrt(1 + x^2) - (4 -x) / 'sqrt(16 + (4-x)^2) Setting equal to 0 and solving: x('sqrt(16 + (4-x)^2) = (4-x)('sqrt(1 + x^2) Squaring: x^2(16 + (4-x)^2) = (4-x)^2(1+x^2) x^2(32 - 8x + x^2) = (16 -8x + x^2)(1+x^2) 32x^2 -8x^3 + x^4 = x^4 - 8x^3 + 17x^2 -8x + 16 Subtracting and adding common elements: 15x^2 = -8x + 16 Setting like a quadratic and solving: 0 = -15x^2 -8x +16 x = .8 (Since having a negative length would be meaningless) Therefore the total length is: T = 'sqrt(1 + .8^2) + 'sqrt(16 + (4 - .8)^2 = 6.4 miles.
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22:51:35 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?^3 + 32?^2 = x^4 - 8?^3 + 17?^2 - 8? + 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 --> Whew. That was a long one.
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23:04:38 Query 4.6.16. (was problem 12 p 280 ) family a cosh(x/a), a>0. Describe your graphs.
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RESPONSE --> Not 4th edition 4.6.16. In fact, I can't even find it. I assume this question means to describe the family of cosh graphs with varying inputs of a, all positive. So here goes: The cosh function is a hyperbola, meaning it is concave down and is reflected on the y-axis. By dividing the x variable by a, the width of the hyberbola is changed, but the value of the y-intercept is the same.
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23:09:21 ** 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 --> Ah, I forgot the a variable in front of the cosh(x/a) function. That would have helped! I also forgot to include the ideas of differentiability and continuity.
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23:12:41 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Some of those modeling questions were difficult! Ah, real life equations. Still, it is applies the techniques learned earlier in useful ways, and is definitely interesting.
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