Homework help...

[KI4MVP]

Thanks drew, I ended up getting those.. just kicked a fucking shit ton of ass on this midterm except for one problem.. couldnt for the life of me figure it out..


Assume the following equation depicting the relationship between price and quantity demanded (x).

p= 100e^-.05x

What is the marginal cost if x=5?

Now. I spent 90% of my time on this question and still couldnt get an answer I felt comfy with. Firstly, youre supposed to find the derivative of the equation. (this is my first attempt). So that is basically 100e^-.05x (-.05). Thats the equation to find the derivative of an E equation.

I also tried flipping in 5 for the x term. This provided me with the closest answer available... it was like, 77.78 and the closest answer given was 77.68. Im sure this isnt the correct answer though.

What the hell am I missing??? Then, last idea, is that p(x) = R and youre supposed to find marginal revenue of that, which would be its' derivative. so this equals

100e^-.05x (-.05) times x =

i dint even fucking know what that answer is. That, theoretically, should be the answer. So what, 100e^-.05x^2 (-.05x^2) = 77.78 * (-.1x) which equals 77.78 * (-.5)? this is clearly not the right fucking answer. I.. i fucking give up

I'm not sure you're setting it up right because the answer to doing it the way you said (i.e. evaluate "derivative of 100e^(-.05x) where x=5") is -3.894

http://www.wolframalpha.com/input/?i=derivative+of+100e^(-.05x)+where+x=5

 

[Shakalu M.D.]

the thing that throws people off is it's teh fact that who knows what and when that changes the odds. If you pick the door in your head and keep it a secret, and then are shown a door that could be your secret door, then on the times your door isn't picked, it doesn't matter if you switch.

If hte person showing the door doesn't know where the car is, so will show the car 1/3 of the time, it doesn't matter if you switch.

It's only when the door you pick is known and an empty door has to be shown by someone who knows both your pick and the car location that it matters.

I spent hours trying to convince my father, who was pretty smart, of the correct answer. I even wrote a computer simulation (in hypercard) to prove it to him. Idon't think he ever quite believed it because he couldn't get past the fact that picking after the door is shown makes it 50-50.

Oh, and for the post above, the person who put this in Parade magazine wasn't a savant, her name was "vos Savant". SHe is the official world record holder for highest IQ. In all of the years she ran the column, she only gave the wrong answer once. That's why this problem became so talked about, when math experts where claiming she was wrong when she actually wasn't.

Actually her IQ score was beaten a few years ago. Guinness has stopped recognizing this category because of how IQ tests can vary.

 

[David.]

kI, positive im setting it up correctly.. I asked the other 500 people in the math class and finally got the right break down..

What I did was multiply an x to the p(x) equation because that's what you do to make it an r equation... So I got 100xe^-.05x then you take the derivative which then you use the product rule (100)(e^-.05x)+(100x)(e^-.05x)(-.05), then you plug in 5 for x I believe and you have the answer. I'm 90% sure I did it correctly



I knew the x would need to be included into the P equation to get R, which you find the derivative of to get marginal rev, but I just distributed it incorrectly, the 100 before the E really threw me off

 

[KI4MVP]

kI, positive im setting it up correctly.. I asked the other 500 people in the math class and finally got the right break down..

What I did was multiply an x to the p(x) equation because that's what you do to make it an r equation... So I got 100xe^-.05x then you take the derivative which then you use the product rule (100)(e^-.05x)+(100x)(e^-.05x)(-.05), then you plug in 5 for x I believe and you have the answer. I'm 90% sure I did it correctly



I knew the x would need to be included into the P equation to get R, which you find the derivative of to get marginal rev, but I just distributed it incorrectly, the 100 before the E really threw me off

this is a different equation than you said in the post I quoted. The x before the e wasn't in the other post.

I get 75e^-0.25 or 58.4101 for that, which wasn't one of the answers you listed unless you typed it wrong.

http://www.wolframalpha.com/input/?i=d/dx+(100+x)/e^(0.05+x)+where+x+=5

Wolfram alpha will solve the equation correctly (when you get the parenthesis right), and you did do the derivative right.

 

[David.]

the x after the e was inserted via product rule im pretty sure, it wasnt in the initial equation.

i use w.a. all the time, me gusta

 

[David.]

The cost function for producing x items is C(x) = 5x^2+125

What is the minimum average cost?

Ok.. I divide the above by x to get the average cost, then find the derivative and set it to 0 to get minimum average cost... this left me with a final equation of (when set to 0)

5-125/x^2 which, in my book, equals 5... this is apparently incorrect...


Secondly.. Im lost as fuck on this one..

2. For the function f(x) = ax^3 +bx^2 , determine a and b so that the point (5,250) is a point of inflection of f(x)

old school rep for answers and explanation, thanks

 

[TopGun]

The cost function for producing x items is C(x) = 5x^2+125

What is the minimum average cost?

Ok.. I divide the above by x to get the average cost, then find the derivative and set it to 0 to get minimum average cost... this left me with a final equation of (when set to 0)

5-125/x^2 which, in my book, equals 5... this is apparently incorrect...


Secondly.. Im lost as fuck on this one..

2. For the function f(x) = ax^3 +bx^2 , determine a and b so that the point (5,250) is a point of inflection of f(x)

old school rep for answers and explanation, thanks

I got the same as you for the first one, seems correct to me.

For the second one, first set the equation equal to 250 when x = 5, so 125a + 25b = 250, simplified to b = 10 - 5a.

Then take the derivative of the original equation, so f'(x) = 3ax^2 + 2bx.

Now set the derivative to 0 when x = 5 and replace b with the equation from before, so 75a + 10(10-5a) = 0. Solve for a and find a = -4.

Then go back to find b, b = 10-5a = 30. Double check the original function and the derivative.

f(x) = -4x^3 + 30x^2
f'(x) = -12x^2 + 60x
When x = 5, f(x) = 250, f'(x) = 0.

 

[natedagg]

The cost function for producing x items is C(x) = 5x^2+125

What is the minimum average cost?

Ok.. I divide the above by x to get the average cost, then find the derivative and set it to 0 to get minimum average cost... this left me with a final equation of (when set to 0)

5-125/x^2 which, in my book, equals 5... this is apparently incorrect...


Secondly.. Im lost as fuck on this one..

2. For the function f(x) = ax^3 +bx^2 , determine a and b so that the point (5,250) is a point of inflection of f(x)

old school rep for answers and explanation, thanks

2nd one is easy - set the 2nd derivative of the function to 0 then plug in 5, 250 to solve for a (b should be nullified to a constant from the 2nd deriv), then plug a into orig equation to solve for b. to lazy to do it.

 

[natedagg]

I got the same as you for the first one, seems correct to me.

For the second one, first set the equation equal to 250 when x = 5, so 125a + 25b = 250, simplified to b = 10 - 5a.

Then take the derivative of the original equation, so f'(x) = 3ax^2 + 2bx.

Now set the derivative to 0 when x = 5 and replace b with the equation from before, so 75a + 10(10-5a) = 0. Solve for a and find a = -4.

Then go back to find b, b = 10-5a = 30. Double check the original function and the derivative.

f(x) = -4x^3 + 30x^2
f'(x) = -12x^2 + 60x
When x = 5, f(x) = 250, f'(x) = 0.

I am pretty sure inflection points are when acceleration = 0 aka f"(x), instead of what you suggested, which is the first derivative f'(x), which is the velocity.

 

[David.]

i agree with both answers, but the computer is not on the same page.. still looking for confirmation/correct answers

 

[natedagg]

The cost function for producing x items is C(x) = 5x^2+125

What is the minimum average cost?

Ok.. I divide the above by x to get the average cost, then find the derivative and set it to 0 to get minimum average cost... this left me with a final equation of (when set to 0)

5-125/x^2 which, in my book, equals 5... this is apparently incorrect...


Secondly.. Im lost as fuck on this one..

2. For the function f(x) = ax^3 +bx^2 , determine a and b so that the point (5,250) is a point of inflection of f(x)

old school rep for answers and explanation, thanks

If the 1st problem's answer is 10, then it's just solving for the 2nd derivative.

 

[MattyFos.]

I still don't understand the door scenario being 2/3 instead of 1/2. Heard the explanation in 21.. read the explanation in here and it still doesn't make sense to me.

 

[TopGun]

I got the same as you for the first one, seems correct to me.

For the second one, first set the equation equal to 250 when x = 5, so 125a + 25b = 250, simplified to b = 10 - 5a.

Then take the derivative of the original equation, so f'(x) = 3ax^2 + 2bx.

Now set the derivative to 0 when x = 5 and replace b with the equation from before, so 75a + 10(10-5a) = 0. Solve for a and find a = -4.

Then go back to find b, b = 10-5a = 30. Double check the original function and the derivative.

f(x) = -4x^3 + 30x^2
f'(x) = -12x^2 + 60x
When x = 5, f(x) = 250, f'(x) = 0.

Nate was right, so instead of the original function and the derivative, use the original function and the second derivative.

So you get f''(x) = 6ax + 2b. Set f''(5) = 0.

30a + 2(10-5a) = 0. a = -1, b = 15.

So f(x) = -x^3 + 15x^2, f''(x) = -6x + 30.

When x = 5, f(x) = 250, f''(x) = 0.

 

[Shakalu M.D.]

I still don't understand the door scenario being 2/3 instead of 1/2. Heard the explanation in 21.. read the explanation in here and it still doesn't make sense to me.

I spent a good 30 minutes trying to explain it to my mother, she didn't really get it either. Just try googling the problem, they may have a better explanation.

 

[TopGun]

I still don't understand the door scenario being 2/3 instead of 1/2. Heard the explanation in 21.. read the explanation in here and it still doesn't make sense to me.

I think the key is understanding that going from three doors to two doesn't affect the probability of the choice you already made.

Imagine the same game except instead of three doors, there are one hundred. You select a door with a 1% chance of choosing the correct one. Now, 98 of the incorrect doors are removed, and you are down to the door you selected and one random door from the group. Now, when you chose your door, you had a 1% chance of picking the right one. With two doors left, your door still only has that 1% chance, and the other door has a 99% chance of being the correct door.