#$&* course Mth 271 11/24 730pcomments inserted on last problem uery23
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Given Solution: `a 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t understand why you are testing (x-2) and (x-2)^3 instead of the entire derivative of 4x^3-32. Also I don’t understand why x= -2 is the relative minimum instead of x=2. x=2 would follow the examples in the book @& Your solution is good. The corrected 'given solution' should read 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 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, 4 x^3 - 32 is negative. For x > 2, 4 x^3 - 32 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 -108. That is the function has its minimum at (2, -108). ** *@ ------------------------------------------------ ------------------------------------------------ Self-critique Rating:2 ********************************************* ********************************************* Question: `q problem 2 d 7th edition 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4+4x^-1+4x^-2 f’(x) = -4x^-2-8x^-3 [(-4/x^2) -(8/x^3)]*x^3 = -4x-8 -4x-8 = 0 x = -2 critical # 4(1+ -.25+ .0625) =4(.8125) = 3.25 minimum 4( 1+ (⅕) + (⅕^2)) = 4.96 maximum confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I couldn’t find examples in the book where the values for the endpoints were not the absolute extrema, so I didn’t know to go a step further. I found a reference to global behavior, but not a global minimum. @& The idea is pretty straightforward: You find the relative maxima and minima within the given interval, then you find the endpoint values. If an endpoint value is greater than the greatest relative max, then it is the global max, and if it's less than the least relative min, it's the global min. *@ ------------------------------------------------ ------------------------------------------------ Self-critique Rating:1 ********************************************* ********************************************* Question: `q problem 4 7th edition 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 $4. Price for maximum profit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x= k/p^3 8= k/10^3 k = 8000 x = 8000/p^3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Why wouldn’t the cube root of 8000/x^3 be 20/x ? @& x = 8000 / p^3, not p = 8000 / x^3. So there is no reason to do anything with the expression 8000 / x^3, which doesn't arise in this model. You can rearrange x = 8000 / p^3 to get p^3 = 8000 / x, so that p = 20 / x^(1/3). ????This (copied and pasted from above) is where I got lost: 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 From here when I solved for p in terms of x from x= 8000/p^3, I got p^3 = 8000/x , not p^3 = 8000 x^3. Taking the cube root of 8000/x, I can easily see where 20/x^⅓ comes from and the rest of it makes sense.?????? *@ It would be clearer not to abbreviate initial cost. I took that as a typo of unit, which didn’t make sense since the very next thing was cost per unit. @& The abbreviated problem statements in the queries are for identifying the problem, and cannot be taken as full statements of the problem. The full statement is in the document containing the problems. ????I am working out of the 9th edition, not the 7th. I try to look for the same problem in the 9th edition, but it is not always there. If I had known I needed the 7th edition, I would have got that one.???? *@ I have struggled with this problem for a few days now. I can’t seem to get past question 1 above. I need clarification on that before I attempt to work through the rest of it. ------------------------------------------------ ------------------------------------------------ Self-critique Rating:0 @& Hopefully between my notes and the file I sent you you'll get the information you need to progress. An overview of the solution: demand 8000 / p^3 revenue = price * demand = p * (8000 / p^3) profit = revenue - cost You maximize the profit function by taking its derivative with respect to p, setting it equal to zero, etc.. *@ "" &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ ------------------------------------------------ Self-critique rating: "" &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ ------------------------------------------------ Self-critique rating: #*&! @& Hopefully my notes will help you move forward. More questions are, as always, welcome. *@ "