If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
023. `query 23
Question: `q problem 1 d 7th edition 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.
Your solution:
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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 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). **
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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?
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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. **
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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?
Your solution:
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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 **
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