#$&* course Mth 272 8/2 5:49pm 029.
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Given Solution: `a The x derivative is 2x; at (-2,1,3) we have x = -2 so the slope is 2 * -2 = -4. The slope in the y direction is the y partial derivaitve -2y; at y = 1 this is -2 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat is the slope in the x direction at the given point? Describe specifically how you obtained your result. -4; I got the result by taking the derivative of x^2 which is 2x and plugged the x value (-2) to get -4 ********************************************* Question: `qQuery problem 7.4.65 (was 7.4.61) all second partials of ln(x-y) at (2,1) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: fx = der (ln(x - y)) = 1 / (x-y) = (x - y)^-1 fxx = (x - y)^-1 * -1(x - y)^-2 = -1(x - y)^-2 = -1 / (x - y)^2 fxy = (x - y)^-1 * -1(x - y)^-2 = 1(x - y)^-2 fy = (x - y)^-1 * 1 / (x - y)^-2 = -(x - y)^2 fyy = -(x - y)^-1 * 1 / (x - y)^2 = 1 / -(x - y)^2 Plugging (2, 1) into each yields: fx = 1, fxx = -1, fxy = 1, fy = -1, fyy = 1, fyx = -1 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The first x derivative is found by the Chain Rule to be (x-y)' * 1/(x-y), where the ' is derivative with respect to x. We get fx = 1 * 1 / (x-y) = 1 / (x-y), or if you prefer (x-y)^-1, where fx means the first x derivative. The x derivative of this expression is the derivative of (x-y)^-1, which by the Chain Rule is fxx = (x-y)' * -1 (x-y)^-2 = 1 * -1 * (x-y)^-2 = -1/(x-y)^2; here fxx means second x derivative and the ' means derivative with respect to x. fxy is the y derivative of fx, or the y derivative of (x-y)^-1, which by the Chain Rule is fxy = (x-y)' * -1 (x-y)^-2 = -1 * -1 * (x-y)^-2 = 1/(x-y)^2; here fxy means the y derivative of the x derivative and the ' means derivative with respect to y. The first y derivative is found by the Chain Rule to be (x-y)' * 1/(x-y), where the ' is derivative with respect to y. We get fy = -1 * 1 / (x-y) = -1 / (x-y), or if you prefer -(x-y)^-1, where fy means the first y derivative. The y derivative of this expression is the derivative of -(x-y)^-1, which by the Chain Rule is fyy = -(x-y)' * -1 (x-y)^-2 = -[1 * -1 * (x-y)^-2] = 1/(x-y)^2; here fyy means second y derivative and the ' means derivative with respect to y. fyx is the x derivative of fy, or the x derivative of -(x-y)^-1, which by the Chain Rule is fyx = -(x-y)' * -1 (x-y)^-2 = -[1 * -1 * (x-y)^-2] = 1/(x-y)^2; here fyx means the x derivative of the y derivative and the ' means derivative with respect to x. When evaluated at (2, 1) the denominator (x - y)^2 is 1 for every second partial. So we easily obtain fxx = -1 fyy = -1 fxy = fyx = +1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 7.4.68 R = 200 x1 + 200 x2 - 4x1^2 - 8 x1 x2 - 4 x2^2; R is revenue, x1 and x2 production of plant 1 and plant 2 ********************************************* Question: `qWhat is the marginal revenue for plant 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x1 is the variable of interest so: R = 200x1 + 200x2 - 4x1^2 - 8x1x2 - 4x2^2 Partial derivatives of x1 are used so: 200 - 8x1 - 8x2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The derivative of R with respect to x1 is 200 + 0 - 4 (2 x1) - 8 x2 - 0; All all derivatives treat x1 as the variable, x2 as constant. Derivatives of 200 x2 and -4 x2^2 do not involve x1 so are constant with respect to x1, hence are zero. So the marginal revenue with respect to plant 1 is 200 - 8 x1 - 8 x2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat is the marginal revenue for plant 2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the same strategy as x1, you end up with 200 - 8x1 - 8x2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The derivative of R with respect to x2 is 0 + 200 - 0 - 8 x1 - 4 ( 2 x2) = 200 - 8 x1 - 8 x2; All all derivatives treat x2 as the variable, x1 as constant. So the marginal revenue with respect to plant 2 is 200 - 8 x1 - 8 x2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhy should the marginal revenue for plant 1 be the partial derivative of R with respect to x1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x1 represents change in plant production, and thus there must be a derivative with respect to x2 to indicate it. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Marginal revenue is the rate at which revenue changes per unit of increased production. The increased production at plant 1 is the change in x1, so we use the derivative with respect to x1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhy, in real-world terms, might the marginal revenue for each plant depend upon the production of the other plant? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the plants were part of a supply and manufacturing chain, plant x1 could be in charge of manufacturing, while x2 could be specialized in logistics. If x1 had low productivity, then x2 will suffer the same fate because they will have as much as x1 has produced. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The marginal revenues for each plant may depend on the each other for a variety of reasons; for example if one plant awaits shipment of a part from the other, or if one plant is somewhat slow resulting in a bottleneck. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat is is about the function that ensures that the marginal revenue for each plant will depend on the production of both plants? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x1 and x2 show up in the original function and also noting using partial derivatives yield the same equation regardless of what variable was used. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The specific reason is that both derivatives contain x1 and x2 terms, so both marginal revenues depend on both the production of plant 1 and of plant 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!