I will need to go back and study these sections more extensively before taking test#3. "ҽގoѠ assignment #023 ݺӯD Applied Calculus I 08-05-2006
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09:09:51 3.2.12 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 f'(x)= 4x^3 -32 set to 0 4x^3 -32=0 x=2 -inf
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09:10:09 ** 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 --> got it.
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09:34:41 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 --> f(x)= 4(1+1/x+1/x^2) f ' (x)= 4[(-x)^ -2+(-2x)^ -3]= -4/x^2 + -8/x^3 -4/x^2 + -8/x^3 = 0 -4/x^2=8/x^3 1/x^2=-2/x^3 x^2= -x^3/2 x^2 + x^3/2 =0 x= -2
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09:38:09 ** 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 did not complete the problem.
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09:53:09 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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09:56:27 ** 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 need to better understand profit function.
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xUzRym assignment #023 ݺӯD Applied Calculus I 08-05-2006
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09:57:52 3.2.12 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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09:57:54 ** 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 -->
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ō|͈|华x_ assignment #024 ݺӯD Applied Calculus I 08-05-2006
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11:06:02 **** Query 3.3.6 picture of y = x^5 + 5x^4 - 40x^2; zeros around 0, 2.3, crit around 0, 1.8 **** On which intervals is the function concave upward and of which concave downward? **** On which intervals is the second derivative negative, on which positive?
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RESPONSE --> y = x^5 + 5x^4 - 40x^2 y ' = 4x^4 + 20x^3 -80x y '' = 16x^3 + 60x^2 -80 -inf,0 test point -1= -36 less than 0 conc dnwrd 0,1.8 test point 1= -4 less than 0 conc dnwrd 1.8,2.3 test point 2= 288 >0 conc upward
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11:07:07 ** A function is concave upward when the second derivative is positive, concave downward when the second derivative is negative. If y = x^5 + 5x^4 - 40x^2 then y' = 5 x^4 + 20 x^3 - 80 x so y'' = 20 x^3 + 60 x^2 - 80. y'' = 0 when 20 x^3 + 60 x^2 - 80 = 0, or dividing this equation by 20 when x^3 + 3 x^2 - 4 = 0. We easily find the solution x = 1 by trial and error, just substituting simple integers. Then we can divide by x-1 to get x^2 + 4 x + 4, which factors to give us (x+2)^2. Thus y'' = (x-1)(x+2)^2. So y'' = 0 when x = 1 or x = -2. y'' can therefore change signs only at x = 1 or at x = -2. However the nature of the zero at x = -2 is parabolic so y'' doesn't change sign at this point. The only sign change is at x = -1. }For large negative x we have y'' < 0, so y'' < 0 on (infinity, -1), and is positive on (1, infinity). The first derivative is y ' = 5x^4 + 20x^3 - 80x, which is zero when x = 0 and when x = 1.679 approx.. The second derivative is y '' = 20x^3 + 60x^2 - 80. Setting this equal to zero we obtain 20x^3 + 60x^2 - 80 = 0 with solutions x = -2 and x = 1. The second derivative is negative for x < -2, again negative for -2 < x < 1 and positive for x > 1, so the function is concave down for all x < 1 except for the x = -2 point, and concave up for all x > 1. The critical point at x = 0 therefore yields a maximum and the critical point at 1.679 yields a minimum. DER
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11:17:22 **** Query 3.3.19-22 concave up increasing, concave down increasing, concave down decreasing, concave up increasing **** In order list the sign of the first derivatives of the functions represented by the four graphs, and the same for the second derivatives.
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RESPONSE --> cc up inc f ' is pos cc dn inc f ' is pos cc dn dec f ' is neg cc up dec f ' is neg cc up inc f "" is pos cc dn inc f "" is pos cc dn dec f "" is neg cc up dec f "" is neg
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11:18:06 ** First derivative is positive for an increasing function, negative for a decreasing function. So we have for the first derivative: 19. positive 20. positive 21. negative 22. negative Second derivative is positive if concave up, negative if concave down. So we have for the second derivative: 19. positive 20. negative 21. negative 22. positive **
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RESPONSE --> i missed the second derivative.
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11:34:59 **** Query 3.3.30 points of inflection for (1-t)(t-4)(t^2-4) **** List the points of inflection of the graph of the given function and explain how you obtained each.
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RESPONSE --> y=(1-t)(t-4)(t^2-4) y ' = (1-t)(t-4)(2t) + (t-4)(t^2-4)(-1) +(1-t)(t^2-4)(1) = - 5t^2 -2t +12 y"" = -10t -2 point of inflection = 2/5 or .4 take the second der of function, when 3rd der = 0 point of inflection occurs.
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11:36:24 ** A point of inflection is a point where the concavity changes. Since the sign of the second derivative determines concavity, a point of inflection is a point where the sign of the second derivative changes. For a continuous function (the case here since we have a polynomial) the sign of the second derivative changes when the second derivative passes through 0. The function is f = -t^4+5t^3-20t-16 Derivative is f ' = -4t^3+15t^2-20 so f '' = -12t^2+30t, which is a quadratic function with zeros at t=0 or 5/2 . The graph of y'' vs. x is therefore a parabola. For large negative x we have y'' negative, since the leading term is -30 t^2. So on (infinity, 0) we see that y'' will be negative. y'' changes sign at x=0 so on (0, 5/2) we see that y'' is positive, and t =0 is an inflection point. y'' again changes sign at x=5/2 so on (5/2, infinity) we see that y'' is negative and t = 5/2 is an inflection point. COMMON ERROR:Common error by student: f'(t)=(-1)(1)(2t) f''(t)=-2 (- infinity, + infinity) the point of inflection doesn't exist INSTRUCTOR CORRECTION: Your derivative was based on the incorrect idea that (f g ) ' = f ' * g '. Be sure you understand how how fell into this error. You would need to apply the product rule to this function twice. The easier alternative is to multiply it out and take the derivative of the resulting polynomial. **
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12:54:06 **** Query 3.3.50 production level to minimize average cost per unit for cost function C = .002 x^3 + 20 x + 500 **** What is the production level to minimize the average cost and how did you obtain it?
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RESPONSE --> C = .002 x^3 + 20 x + 500 C ' =.006x^2 + 20 C "" = .012x
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12:55:22 ** Ave cost per unit is cost / # of units = C / x, so average cost per unit = (.002x^3 + 20 x + 500)/x = .002 x^2 + 20 + 500/x. derivative of average cost per unit = .004 x - (500/ x^2) Critical numbers occur when derivative is 0: 0 = .004 x - (500/x^2) 500 / x^2 = .004 x so 500 = .004 x^3 and x^3 = 500 / .004 = 125,000. x = 50 (critical number) The second derivative is .004 + (1000 / x^3). For all x > 0 the second derivative is therefore positive. So for x > 0 the graph is concave up, and this shows that the critical point at x = 50 is a minimum. Alternatively you could show that .004 x - (500 / x^2) changes sign at x = 50. COMMON ERROR: Students commonly make the error of minimizing only the given function. Note that you aren't supposed to minimize the cost, but the cost per unit. This is C / x = .002 x^2 + 20 + 500/x. (C/x)' = .004 x - 500 / x^2, which is zero when .004 x - 500 / x^2 = 0; multiplying by x^2 we get .004 x^3 - 500 = 0 so x^3 = 500 / .004 = 62500. Either a first- or second-degree test shows this to be a minimum. **
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RESPONSE --> i wasn't even close need to examine the correct answer.
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۳RYŬK҇믑ۤzU{ assignment #024 ݺӯD Applied Calculus I 08-05-2006
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12:57:05 **** Query 3.3.6 picture of y = x^5 + 5x^4 - 40x^2; zeros around 0, 2.3, crit around 0, 1.8 **** On which intervals is the function concave upward and of which concave downward? **** On which intervals is the second derivative negative, on which positive?
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RESPONSE -->
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12:57:08 ** A function is concave upward when the second derivative is positive, concave downward when the second derivative is negative. If y = x^5 + 5x^4 - 40x^2 then y' = 5 x^4 + 20 x^3 - 80 x so y'' = 20 x^3 + 60 x^2 - 80. y'' = 0 when 20 x^3 + 60 x^2 - 80 = 0, or dividing this equation by 20 when x^3 + 3 x^2 - 4 = 0. We easily find the solution x = 1 by trial and error, just substituting simple integers. Then we can divide by x-1 to get x^2 + 4 x + 4, which factors to give us (x+2)^2. Thus y'' = (x-1)(x+2)^2. So y'' = 0 when x = 1 or x = -2. y'' can therefore change signs only at x = 1 or at x = -2. However the nature of the zero at x = -2 is parabolic so y'' doesn't change sign at this point. The only sign change is at x = -1. }For large negative x we have y'' < 0, so y'' < 0 on (infinity, -1), and is positive on (1, infinity). The first derivative is y ' = 5x^4 + 20x^3 - 80x, which is zero when x = 0 and when x = 1.679 approx.. The second derivative is y '' = 20x^3 + 60x^2 - 80. Setting this equal to zero we obtain 20x^3 + 60x^2 - 80 = 0 with solutions x = -2 and x = 1. The second derivative is negative for x < -2, again negative for -2 < x < 1 and positive for x > 1, so the function is concave down for all x < 1 except for the x = -2 point, and concave up for all x > 1. The critical point at x = 0 therefore yields a maximum and the critical point at 1.679 yields a minimum. DER
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12:57:12 **** Query 3.3.19-22 concave up increasing, concave down increasing, concave down decreasing, concave up increasing **** In order list the sign of the first derivatives of the functions represented by the four graphs, and the same for the second derivatives.
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12:57:36 ** First derivative is positive for an increasing function, negative for a decreasing function. So we have for the first derivative: 19. positive 20. positive 21. negative 22. negative Second derivative is positive if concave up, negative if concave down. So we have for the second derivative: 19. positive 20. negative 21. negative 22. positive **
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12:57:42 **** Query 3.3.30 points of inflection for (1-t)(t-4)(t^2-4) **** List the points of inflection of the graph of the given function and explain how you obtained each.
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12:57:45 ** A point of inflection is a point where the concavity changes. Since the sign of the second derivative determines concavity, a point of inflection is a point where the sign of the second derivative changes. For a continuous function (the case here since we have a polynomial) the sign of the second derivative changes when the second derivative passes through 0. The function is f = -t^4+5t^3-20t-16 Derivative is f ' = -4t^3+15t^2-20 so f '' = -12t^2+30t, which is a quadratic function with zeros at t=0 or 5/2 . The graph of y'' vs. x is therefore a parabola. For large negative x we have y'' negative, since the leading term is -30 t^2. So on (infinity, 0) we see that y'' will be negative. y'' changes sign at x=0 so on (0, 5/2) we see that y'' is positive, and t =0 is an inflection point. y'' again changes sign at x=5/2 so on (5/2, infinity) we see that y'' is negative and t = 5/2 is an inflection point. COMMON ERROR:Common error by student: f'(t)=(-1)(1)(2t) f''(t)=-2 (- infinity, + infinity) the point of inflection doesn't exist INSTRUCTOR CORRECTION: Your derivative was based on the incorrect idea that (f g ) ' = f ' * g '. Be sure you understand how how fell into this error. You would need to apply the product rule to this function twice. The easier alternative is to multiply it out and take the derivative of the resulting polynomial. **
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12:57:47 **** Query 3.3.50 production level to minimize average cost per unit for cost function C = .002 x^3 + 20 x + 500 **** What is the production level to minimize the average cost and how did you obtain it?
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as}cn assignment #025 ݺӯD Applied Calculus I 08-05-2006
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13:29:26 3.4.6 find two positive numbers such that the product is 192 and a sum of the first plus three times the second is a minimum What are the two desired numbers and how did you find them?
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RESPONSE --> min= m 192=y+x y=192/x m= y+3x =192/x +3x m' = - 192/x^2 +3 0 = - 192/x^2 +3 192/3=x^2 x=8 critical# min (8,24)
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13:30:37 ** First set up the primary equation S=x+3y (y being the 2nd number) and the secondary equation xy=192. So S = x + 3(192/x). We now maximize the function by finding critical points (points where the derivative is zero) and testing to see whether each gives a max, a min, or neither. S ' = 1 - 576 / x^2, which is zero when x = sqrt(576) = 24 (or -24, but the problem asks for positive numbers). For this value of x we get y = 192 / x = 192 / 24 = 8. So the numbers are x = 24 and y = 8. }Note that x = 24 does result in a min by the first derivative test, since S ' is negative for x < 24 and positive for x > 24. **
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RESPONSE --> i got it right, a rare occasion.
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13:37:43 3.4.18 80 apple trees yield and average of 400 per tree; each additional tree decreases the yield by 32 apples per tree. Maximize yield.
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bī assignment #025 ݺӯD Applied Calculus I 08-05-2006
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13:39:41 3.4.6 find two positive numbers such that the product is 192 and a sum of the first plus three times the second is a minimum What are the two desired numbers and how did you find them?
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13:40:08 ** First set up the primary equation S=x+3y (y being the 2nd number) and the secondary equation xy=192. So S = x + 3(192/x). We now maximize the function by finding critical points (points where the derivative is zero) and testing to see whether each gives a max, a min, or neither. S ' = 1 - 576 / x^2, which is zero when x = sqrt(576) = 24 (or -24, but the problem asks for positive numbers). For this value of x we get y = 192 / x = 192 / 24 = 8. So the numbers are x = 24 and y = 8. }Note that x = 24 does result in a min by the first derivative test, since S ' is negative for x < 24 and positive for x > 24. **
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13:41:11 3.4.18 80 apple trees yield and average of 400 per tree; each additional tree decreases the yield by 32 apples per tree. Maximize yield.
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13:41:55 How many trees should be planted and what will be the maximum yield?
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13:42:04
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13:43:47 ** If we let x stand for the number of trees added to the 80 then the yield per tree is 400 - 32 x, and there would be 80 + x trees. The total yield would therefore be total yield = number of trees * yield per tree = (80 + x) * ( 400 - 32 x) = -32 x^2 + -2160 x + 32000, which is maximized when x = -34 approx.; this indicates -34 trees in addition to the 80, or 46 trees total. Another approach is to assume that only the additional trees experience the decrease. However it doesn't make sense for the yield decrease to apply only to the added trees and not to the original 400. If you're gonna crowd the orchard every tree should suffer. In any case, if we make the unrealistic assumption that the original 80 trees maintain their 400-apple-per-tree yield, and that the x additional trees each have a yield of 32 x below the 400, we have x added trees each producing 400 - 32 x apples, so we produce x (400 - 32 x) = 400 x - 32 x^2 additional apples. We therefore maximize the expression y = 400 x - 32 x^2. We obtain y ' = 400 - 64 x, which is 0 when -64 x + 400 = 0 or x = 6.25. Since y ' is positive for x < 6.25 and negative for x > 6.25 we see that 6.25 will be our maximizing value. We can't plant 6.25 trees, so the actual maximum must occur for either 6 or 7 trees. We easily see that the max occurs for 6 additional trees. So according to this interpretation we plant 86 trees. **
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RESPONSE --> i could not come up with an equation
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13:43:51 Add comments on any surprises or insights you experienced as a result of this assignment.
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_]×ܩx߳r assignment #026 ݺӯD Applied Calculus I 08-05-2006
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14:26:22 **** Query 3.5.12 find the price per unit p for maximum profit P if C = 35x+500, p=50-.1`sqrt(x) **** What price per unit produces the maximum profit?
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RESPONSE --> P=R-C =x[50-0.1(x)^1/2] - 3x+500 = 47x - 0.1(x)^3/2 +500 P ' = 47- 0.15(x)^-1/2 = 47 - 0.15/x^1/2 47 - 0.15/x^1/2= 0 - 0.15/x^1/2= -47 x^1/2 = 0.15/47 x= 10 x 10^ -6 50-.1`sqrt(10 x 10^ -6) $ 50.0003
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14:27:37 ** Revenus is price * number sold: R = xp. Since p = 50 - .1 sqrt(x) we have R = x(50 - .1 `sqrt (x)) = 50x - .1x^(3/2) Price is revenue - cost: P = R - C = 50x - .1 x^(3/2) - 35x - 500. Simplifying: P = 15x - .1x^(3/2) - 500 Derivative of profit P is P ' = 15 -.15 x^(1/2). Derivative is zero when 15 - .15 x^(1/2) = 0; solving we get x = 10,000. 2d derivative is .075 x^-(1/2), which is negative, implying that x = 10000 gives a max. When x = 10,000 we get price p = 50 - .1 sqrt(x) = 50 - .1 * sqrt(10,000) = 40. Price is $40. **
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RESPONSE --> i got off track when i solved the der = 0 wrong
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14:27:48 **** Query 3.5.22 amount deposited proportional to square of interest rate; bank can reinvest at 12%. What interest rate maximizes the bank's profit? **** What is the desired interest rate?
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14:27:51 ** According to my note here amount deposited A is proportional to the square of interest rate r so A = k r^2 for some proportionality constant k. The interest paid at rate r on amount A is A * r. The bank can reinvest at 12% so it gets return A * .12. The bank therefore nets .12 * A - r * A = (.12 - r) * A. Since A = k r^2 the bank nets profit P = (.12 - r) * (k r^2) = k * (.12 r^2 - r^3). We maximize this expression with respect to r: dP/dr = k * (.24 r - 3 r^2). dP/dr = 0 when .24 r - 3 r^2 = 0, when 3 r ( .08 - r) = 0, i.e., when r = 0 or r = .08. The second derivative is -6 r + .24, which is negative for r > .06. This shows that the critical point at r = .08 is a maximum. The max profit is thus P = (.12 * .08 - .08^3) * k = (.096 - .0016) k = .080 k. In order to find the optimal interest rate it is not necessary to find the proportionality constant k. However if the proportionality constant was known we could find the max profit. **
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ÈnYx assignment #026 ݺӯD Applied Calculus I 08-05-2006
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14:28:27 **** Query 3.5.12 find the price per unit p for maximum profit P if C = 35x+500, p=50-.1`sqrt(x) **** What price per unit produces the maximum profit?
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14:28:29 ** Revenus is price * number sold: R = xp. Since p = 50 - .1 sqrt(x) we have R = x(50 - .1 `sqrt (x)) = 50x - .1x^(3/2) Price is revenue - cost: P = R - C = 50x - .1 x^(3/2) - 35x - 500. Simplifying: P = 15x - .1x^(3/2) - 500 Derivative of profit P is P ' = 15 -.15 x^(1/2). Derivative is zero when 15 - .15 x^(1/2) = 0; solving we get x = 10,000. 2d derivative is .075 x^-(1/2), which is negative, implying that x = 10000 gives a max. When x = 10,000 we get price p = 50 - .1 sqrt(x) = 50 - .1 * sqrt(10,000) = 40. Price is $40. **
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14:45:45 **** Query 3.5.22 amount deposited proportional to square of interest rate; bank can reinvest at 12%. What interest rate maximizes the bank's profit? **** What is the desired interest rate?
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RESPONSE --> amount dep/y = (int rate)^2/x y= amount dep [x/ (int rate)^2] = amount dep [ (x)(int rate)^ - 2 = amount dep[ -2x (int rate)^ -3 + (int rate)^ -2
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14:46:38 ** According to my note here amount deposited A is proportional to the square of interest rate r so A = k r^2 for some proportionality constant k. The interest paid at rate r on amount A is A * r. The bank can reinvest at 12% so it gets return A * .12. The bank therefore nets .12 * A - r * A = (.12 - r) * A. Since A = k r^2 the bank nets profit P = (.12 - r) * (k r^2) = k * (.12 r^2 - r^3). We maximize this expression with respect to r: dP/dr = k * (.24 r - 3 r^2). dP/dr = 0 when .24 r - 3 r^2 = 0, when 3 r ( .08 - r) = 0, i.e., when r = 0 or r = .08. The second derivative is -6 r + .24, which is negative for r > .06. This shows that the critical point at r = .08 is a maximum. The max profit is thus P = (.12 * .08 - .08^3) * k = (.096 - .0016) k = .080 k. In order to find the optimal interest rate it is not necessary to find the proportionality constant k. However if the proportionality constant was known we could find the max profit. **
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RESPONSE --> no where near close to that one.
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H{z{睒Б}~^Ʌ^m~ assignment #026 ݺӯD Applied Calculus I 08-05-2006
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14:47:03 **** Query 3.5.12 find the price per unit p for maximum profit P if C = 35x+500, p=50-.1`sqrt(x) **** What price per unit produces the maximum profit?
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14:47:06 ** Revenus is price * number sold: R = xp. Since p = 50 - .1 sqrt(x) we have R = x(50 - .1 `sqrt (x)) = 50x - .1x^(3/2) Price is revenue - cost: P = R - C = 50x - .1 x^(3/2) - 35x - 500. Simplifying: P = 15x - .1x^(3/2) - 500 Derivative of profit P is P ' = 15 -.15 x^(1/2). Derivative is zero when 15 - .15 x^(1/2) = 0; solving we get x = 10,000. 2d derivative is .075 x^-(1/2), which is negative, implying that x = 10000 gives a max. When x = 10,000 we get price p = 50 - .1 sqrt(x) = 50 - .1 * sqrt(10,000) = 40. Price is $40. **
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