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course MTH 272 12/15 about 2:45 pm 029..............................................
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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To get the slope in the x direction we must differentiate with respect to x and plug in the values of the point Zx(-2,1,3)=2x-0=2x=2*(-2)=-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=1/(x-y)*1=1/(x-y) Fy=1/(x-y)*-1+-1/(x-y) Fxx=-1*(1/(x-y)^2)*1=-1/(x-y)^2 ->-1/(2-1)^2=-1/1=-1 Fyy=-(-1)*(-1/(x-y)^2)=-1/(x-y)^2 ->-1/(2-1)^2=-1/1=-1 Fxy=-1*(1/(x-y)^2)*-1=1/(x-y)^2 ->1/(2-1)^2=1/1=1 Fyx=-1*(1/(x-y)^2)*-1=1/(x-y)^2 ->1/(2-1)^2=1/1=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: Rx1=200+0-8x1-8x2-0=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: Rx2=0+200-0-8x1-8x2=200-8x1-8x2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The derivative of R with respect to x2 is 0 + 200 - 0 - 8 x1 - 4 ( 2 x2) = 200 - 8 x1 - 8 x2; All all derivatives treat x2 as the variable, x1 as constant. So the marginal revenue with respect to plant 2 is 200 - 8 x1 - 8 x2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhy should the marginal revenue for plant 1 be the partial derivative of R with respect to x1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You are looking at the change in production of plant 1 (‘delta x1) while plant 2 remains constant. confidence rating #$&*: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 one plant is not producing, it could cause delays to the other plant resulting in a decrease in marginal revenue, if plant 1 relies on the production of plant 2. If these were competing plants, if one plant is not producing , the other plant would have an increase in marginal revenue, if supply equals demand. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The marginal revenues for each plant may depend on the each other for a variety of reasons; for example if one plant awaits shipment of a part from the other, or if one plant is somewhat slow resulting in a bottleneck. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat is is about the function that ensures that the marginal revenue for each plant will depend on the production of both plants? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The partial derivates with respect to x1 and x2 both contain both variables. confidence rating #$&*: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: ________________________________________ `gr91 #$&*#$&* course MTH 272 12/15 about 2:45 pm 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To get the slope in the x direction we must differentiate with respect to x and plug in the values of the point Zx(-2,1,3)=2x-0=2x=2*(-2)=-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=1/(x-y)*1=1/(x-y) Fy=1/(x-y)*-1+-1/(x-y) Fxx=-1*(1/(x-y)^2)*1=-1/(x-y)^2 ->-1/(2-1)^2=-1/1=-1 Fyy=-(-1)*(-1/(x-y)^2)=-1/(x-y)^2 ->-1/(2-1)^2=-1/1=-1 Fxy=-1*(1/(x-y)^2)*-1=1/(x-y)^2 ->1/(2-1)^2=1/1=1 Fyx=-1*(1/(x-y)^2)*-1=1/(x-y)^2 ->1/(2-1)^2=1/1=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: Rx1=200+0-8x1-8x2-0=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: Rx2=0+200-0-8x1-8x2=200-8x1-8x2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The derivative of R with respect to x2 is 0 + 200 - 0 - 8 x1 - 4 ( 2 x2) = 200 - 8 x1 - 8 x2; All all derivatives treat x2 as the variable, x1 as constant. So the marginal revenue with respect to plant 2 is 200 - 8 x1 - 8 x2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhy should the marginal revenue for plant 1 be the partial derivative of R with respect to x1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You are looking at the change in production of plant 1 (‘delta x1) while plant 2 remains constant. confidence rating #$&*: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 one plant is not producing, it could cause delays to the other plant resulting in a decrease in marginal revenue, if plant 1 relies on the production of plant 2. If these were competing plants, if one plant is not producing , the other plant would have an increase in marginal revenue, if supply equals demand. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The marginal revenues for each plant may depend on the each other for a variety of reasons; for example if one plant awaits shipment of a part from the other, or if one plant is somewhat slow resulting in a bottleneck. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat is is about the function that ensures that the marginal revenue for each plant will depend on the production of both plants? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The partial derivates with respect to x1 and x2 both contain both variables. confidence rating #$&*: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 ------------------------------------------------