KI4MVP
formerly LJ4MVP
- Joined
- Jun 30, 2005
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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
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