course Mth 271 خcH߷assignment #023
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20:07:36 3.2.10 all relative extrema of x^4 - 32x + 4 Give the x and y coordinates of all the relative extrema of x^4 - 32x + 4.
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RESPONSE --> f(x) = x^4 - 32x + 4 First, I took the derivative of the function. f'(x) = 4x^3 - 32 I set the expression equal to zero and solved for x to find the critical points 4x^3 - 32 = 0 4x^3 = 32 x^3 = 8 x = 2 I then substitute 2 into the original function f'(2) = 2^4 - 32 * 2 + 4 f'(2) = 16 - 64 + 4 = -108 Therefore the coordinates of the relative extremum is (2, -44) confidence assessment: 2
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20:08:48 The procedure is to find the critical numbers, where the derivative is zero, since at a 'peak' or a 'valley' the function levels off and the derivative is for that one instant zero. The derivative of this function is 4 x^3 - 32. 4 x^3 - 32 = 4 ( x^3 - 8) = 4 ( x-2)^3 has a zero at x = 2. This is the only value for which the derivative is zero and hence the only critical point. For x < 2, x - 2 is negative and hence (x-2)^3 is negative. For x > 2, x-2 is positive and hence (x-2)^3 is positive. So the derivative changes from negative to positive at this zero. This means that the function goes from decreasing to increasing at x = -2, so x = -2 is a relative minimum of x^4 - 32x + 4. The value of the function at the relative minimum is -44. That is the function has its minimum at (2, -44). **
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RESPONSE --> I forgot to include the information about the test intervals that told me the point (2, -44) was a relative minimum. self critique assessment: 2
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20:17:18 3.2.30 abs extrema of 4(1+1/x+1/x^2) on [-4,5] What are the absolute extrema of the given function on the interval?
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RESPONSE --> g(x) = 4(1+1/x+1/x^2) g(x) = 4 + 4x^-1 + 4x^-2 g'(x) = -4x^-2 - 8x^-3 Now I set the expression equal to zero and solve for x -4x^-2 - 8x^-3 = 0 -4(x^-2 + 2x^-3) = 0 -4(x^-2)(1 + 2x^-1) = 0 x = 0 cannot be a critical point because it makes the orginal function undefined. 1 + 2x^-1 = 0 2x^-1 = -1 1/x = -1/2 I cross multiply to get x = -2 Now I substitute the critical point -2 and the endpoints from the closed interval [-4, 5] into the original function to find the absolute extrema. f(-4) = 3.25 f(-2) = 3 f(5) = 4.96 I can see that -2 is the absolute minimum and 5 is the absolute maximum. confidence assessment: 3
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20:20:18 the derivative of the function is -4/x^2 - 8 / x^3. Multiplying through by the common denominator x^3 we see that -4/x^2 - 8 / x^3 = 0 when x^3( -4/x^2 - 8 / x^3 ) = 0, x not 0. This simplified to -4 x - 8 = 0, which occurs when x = -2. At x = -2 we have y = 4 ( 1 + 1 / (-2) + 1 / (-2)^2 ) = 4 ( 1 - .5 + .25) = 4(.75) = 3. Thus (-2, 3) is a critical point. Since large negative x yields a negative derivative the derivative for all x < -2 is negative, and since as x -> 0 from 'below' the derivative approaches +infinity the derivative between x=-2 and x = 0 is positive. Thus the derivative goes from negative to positive at x = 2, and the point is a relative minimum. A second-derivative test could also be used to show that the point is a relative minimum. We also need to test the endpoints of the interval for absolute extrema. Testing the endpoints -4 and 5 yields 4(1+1/(-4)+1/(-4)^2) = 3.25 and 4(1+1/5+1/25) = 4(1.24) = 4.96 at x = 5. However these values aren't necessarily the absolute extrema. Recall that the derivative approaches infinity at x = 0. This reminds us to check the graph for vertical asymptotes, and we find that x = 0 is a vertical asymptote of the function. Since as x -> 0 the 1 / x^2 terms dominates, the vertical asymptote will approach positive infinity on both sides of zero, and there is no absolute max; rather the function approaches positive infinity. However the min at (-2, 3), being lower than either endpoint, is the global min for this function. **
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RESPONSE --> I see that there was an asymptote at zero which means that 5 was not the absolute maximum. self critique assessment: 2
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22:25:38 3.2.44 demand x inversely proportional to cube of price p>1; price $10/unit -> demand 8 units; init cost $100, cost per unit $4. Price for maximum profit?
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RESPONSE --> I'm not really sure how to do this problem. Since x is inversely proportional to p^3, I wrote p^3 = k/x p = 10 when x = 8 10^3 = k/8 1000 = k/8 k = 8000 p^3 = 8000/x I then took the cube of both sides p = 20x^(-1/3) I see that cost is 100 plus 4 for each additional unit C = 4x + 100 I know that revenue is price times units sold R = px R = 20x^(-1/3) * x R = 20x^(2/3) I know that profit is equal to revenue minus cost P = R - C P = 20x^(2/3) - (4x + 100) P = 20x^(2/3) - 4x - 100 Since I need the maximum profit with relationship to price, I differentiate with respect to p d/dp[P] = d/dp[20x^(2/3) - 4x - 100 dP/dp = (40/3)x^(-1/3) dx/dp - 4 dx/dp I'm not sure what to do from this point or whether I'm on the right track or not. confidence assessment: 1
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22:36:52 If x is inversely prop to the cube of price, with x = 8 when p =10, then we have: x = k/p^3. Substituting and solving for k: 8 = k / 10^3 8 = k / 1000 k = 8000 So x = 8000/ p^3. We want to maximize profit in terms of x. Profit is revenue - cost and revenue is price * demand = x * p. The demand function is found by solving for p in terms of x: p^3 = 8000/x^3 p = 20/ x^(1/3) The revenue function is therefore R = xp = x (20/ x^(1/3) = 20 x ^ (2/3). The cost function is characterized by init cost $100 and cost per item = $4 so we have C = 100 + 4x The profit function is therefore P = profit = revenue - cost =20x ^(2/3) - 100 - 4x. We want to maximize this function, so we find its critical values: P ' = 40/ 3x^(1/3) - 4 Setting P' = 0 we get 0 = 40/ 3x^(1/3) - 4 4 = 40/ 3x^(1/3) 3x^(1/3) = 40/4 3x^(1/3) = 10 x^(1/3) = 10/3 x = 37.037 units For x < 37.037 we have P ' positive and for x > 37.037 we have P ' negative. So the derivative goes from positive to negative, making x = 37.037 a relative maximum. At the endpoint x = 0 the profit is negative, and as x -> infinity the profit function is dominated by the -4x and becomes negative. At x = 37.037 we find that profit = 20* 37.037^(2/3) - 100 - 4 x profit = -26, approx. This is greater than the endpoint value at x = 0 so this is the maximum profit. This is negative, so we're going to lose money. The graph of the profit function starts at profit -100, peaks at profit -26 when about 37 items are sold, then decreases again. Alternative solution, with demand expressed and maximized in terms of price p: Demand is inversely proportional to cube of price so x = k / p^3. When p = 10, x is 8 so 8 = k / 10^3 and k = 8 * 10^3 = 8000. So the function is x = 8000 / p^3. The cost function is $100 + $4 * x, so the profit is profit = revenue - cost = price * demand - cost = p * 8000 / p^3 - ( 100 + 4 x) = 8000 / p^2 - 100 - 4 ( 8000 / p^3) = -100 + 8000 / p^2 - 32000 / p^3. We maximize this function by finding the derivative -16000 / p^3 + 96000 / p^4 and setting it equal to zero. We obtain -16000 / p^3 + 96000 / p^4 = 0 or -16000 p + 96000 = 0 so p = (96000 / 16000) = 6. For large p the derivative is negative, so the derivative is going from positive to negative and this is a relative max.. We also have to check the endpoint where p = 1. At this price the profit would be -23,900, so the function does have a maximum at p = 6. Note that the above solution in terms of p then gives demand x = 8000 / p^3 = 8000 / 6^3 = 37 approx, which is consistent with the solution we got in terms of x. The revenue would be 6 * 37 = 222, approx.. Cost would be 100 + 4 * 37 = 248 approx, and the profit would be $222-$248=-$26. That is, we're going to lose money, but better to lose the $26 than the $23,900 **
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RESPONSE --> I see that I was on the right track. I also see that with P = 20x^(2/3) - 100 - 4x, I didn't need to diffrentiate with respect to p again. I just needed to take the derivative and find the critical point. I also see that I could have begun with the demand function expressed in terms of p. That way I would have found the price that yielded the most profit instead of the number of units that gave the most profit. self critique assessment: 2
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