open qa 23

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course Mth 271

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.

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Your solution:

F’(x) = 4x^3 - 32

4x^3 - 32 = 0

4x^3 = 32

X^3 = 8

X = 2

2^4 - 32 * 2 + 4 = -44

Function has relative extrema at ( 2, -44)

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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). **

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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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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

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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?

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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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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?

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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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