#$&* course Mth 271 Question: `q problem 1 d 7th edition 3.2.12 all relative extrema of x^4 - 32x + 4
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Given Solution: `a The procedure is to find the critical numbers, where the derivative is zero, since at a 'peak' or a 'valley' the function levels off and the derivative is for that one instant zero. The derivative of this function is 4 x^3 - 32. 4 x^3 - 32 = 4 ( x^3 - 8) = 4 ( x-2)^3 has a zero at x = 2. This is the only value for which the derivative is zero and hence the only critical point. For x < 2, x - 2 is negative and hence (x-2)^3 is negative. For x > 2, x-2 is positive and hence (x-2)^3 is positive. So the derivative changes from negative to positive at this zero. This means that the function goes from decreasing to increasing at x = -2, so x = -2 is a relative minimum of x^4 - 32x + 4. The value of the function at the relative minimum is -44. That is the function has its minimum at (2, -44). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q problem 2 d 7th edition 3.2.30 abs extrema of 4(1+1/x+1/x^2) on [-4,5] What are the absolute extrema of the given function on the interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: G’(x) = 4 (0 + (-1/x^2) + (-2x/x^4)) = 4 (-1/x^2 - 2x/x^4) -(x^2 + 2x) = 0 -x(x + 2) =0 X = 0 and x = -2 X = 0 G(0) = 4 X = -2, g(-2) = = 3 G(-4) = 1/16 G(5) = 124/25 = 4.96 Function has absolute minimum at 1/16 and absolute maximum at 124/25 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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 $4. Price for maximum profit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* 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 $4. Price for maximum profit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: #*&!