#$&* course Mth 173 028. `query 28
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Given Solution: `aIf 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.. So the fence is about 65 ft in the x direction, 46 ft in the other. The cost associated with the solution x = 65 is C(65) = 35 * 65 + 150 000 / 65 = $4582.69 To double-check this: The fourth side has length 65 ft. At $10/ft it will cost $650 to fence it. The other three sides have total length 65 ft + 46 ft + 46 ft = 157 ft. At $25/foot it will cost $25 * 157 = $3925. The total cost is therefore $4575. The 65 ft result is approximate, so we don't expect an exact result. $4575 is within a small fraction of a percent of $4582.69, so the two results appear to agree. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qQuery 4.5.21 (was problem 18 p 276 ) towns at (0,-1) and (4,-4), river on x axis; find min dist of pipe to supply both. What is the minimum total length of pipe? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First town: sqrt(1^2 + x^2) Second town: sqrt(4-x)^2 + 4^2) L ' (x) = x / sqrt(1 + x^2) - ( 4 - x) / sqrt(16 + (4-x)^2) after simplifying: 15 x^2 + 8 x - 16 = 0 We get x=.8 and x=-4/3 So L(x) has a mininum at x=.8 Which means there is a minimum of 6.4 miles of pipe confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a If the pipes are run from the point (x, 0) along the river, 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.. The pipe runs from (0, -1) to (.8, 0), then to (4, -4). You can use the Pythagorean Theorem with each of the resulting right triangles to verify that the total length is about 6.4 miles. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qQuery 4.6.16. (was problem 12 p 280 ) family a cosh(x/a), a>0. Describe your graphs. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph is continuous and differentiable. For a positive x, the graph is positive. For a negative x, the graph is negative. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: