course Mth 272 8/1 2 pm 29.
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the slope in the x direction at the given point? Describe specifically how you obtained your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the slope in the x direction, which was -4, I took the first derivative in reference to x and then plugged in the given x-value (2) into the derivative. ********************************************* Question: `qQuery problem 7.4.65 (was 7.4.61) all second partials of ln(x-y) at (2,1) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the derivatives in reference to x and y, we need to use the chain rule. We find that Z’x = (1)(1/(x-y)) = 1/(x-y) and Z’y = (-1)(1/(x-y)) = -1/(x-y). Now we can find the second partial derivatives for x and y. So Z’xx = (1)(-1/(x-y)^2) = -1/(x-y)^2. Z’xy = (-1)(-1/(x-y)^2) = 1/(x-y)^2. Z’yy = (-1)(-1/(x-y)^2) = 1/(x-y)^2. Z’yx = (-1)(-1/(x-y)^2) = 1/(x-y)^2. Now we can evaluate all our second partial derivatives by plugging in the given values (2,1). We find that Z’xx = -1, Z’xy = 1, Z’yy = 1, and Z’yx = 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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: To find the marginal revenue for the plants, we first need to take the derivative in reference to both x’s. When we take the derivative of the equation in reference to x1 according to our derivative rules, we end up with Rx1 = 200 – 8x1 -8x2. When we take the derivative in reference to x2, we get Rx2 = 200 – 8x1 – 8x2. We can now plug in the given values for x1 and x2 to solve for both plants. Since both plants have the same marginal revenue equation, we find that they both equal 200 – 8(4) – 8(12) = 72. 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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the marginal revenue for plant 2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because we found the marginal revenue equation for plant 2 to be Rx2 = 200 – 8x1 – 8x2 in the previous solution, we can plug in the given values of x1 and x2 to find the actual marginal revenue. So 200 – 8(4) – 8(12) = 72. 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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhy should the marginal revenue for plant 1 be the partial derivative of R with respect to x1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because we know that any revenue equation has the marginal revenue equation of its derivative, we know that this revenue equation is no different. Since we are also looking specifically at the revenue of plant 1, we need to take the derivative of the revenue equation with regards to x1 to solve accurately for just that plant. 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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhy, in real-world terms, might the marginal revenue for each plant depend upon the production of the other plant? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The marginal revenue of the two plants might depend on the production of the other because the plants may be making two different parts of the same product, or work to ship products together, or to work to produce various specializations on a product. 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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: Because the original revenue function contains both x1 and x2, when we take the derivative in regards to x1 and x2 individually, we find that there are components of both plants in the individual mariginal revenue equations. 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):