Calculus Help For Osaka!!

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Calculus Help For Osaka!!

Postby Mangafanatic » Tue Sep 05, 2006 7:10 pm

Much to my chagrin, I must take calculus for my selected major, but I'm really struggling with it. Really struggling. So anyways, I'll trudged my way through all but three of the problems and I'm just at a loss for how to solve them, so I was just wondering if any of my Math-tastic CAA friends could give me a hand! I lurve you people. Thanks!

problem 1: The revenue of a charter bus company depends on the number o unsold seats. If the revenue R(x) is given by r(x)=5000+50x-x^2 where is the number of unsold seats, find the maximum revenue and the number of unsold seats that correspond to maximum revenue.

problem 2: The demand for a certain type of cosmetic is given by:

p=500-x

where p is the price in dollars when x units are demanded.

a. find the revenue R(x) that would be obtained at a price (hint: revenue=demandxprice)

b. Graph the revenue function R(x).

c. From the graph of the revenue function, estimate the price that will produce maximum revenue.

d. What is the maximum revenue.

Problem three: According to the recent data from the teachers insurance and annuity association, the survival function gives the probability that an individual who reaches the age of 65 will live atleast x decades (10x) longer.

a. Find the median length of life for people who reach 65, that is, the age for which the survival rate is .50.

b. Find the age beyond which virtually nobody lives.
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Postby Puritan » Tue Sep 05, 2006 7:39 pm

Calculus! I love calculus! (Sadly, I'm not joking) If you're having problem with derivatives and integration, I'd be happy to do a tutorial, but I'm assuming you get the basics. Alright, here it goes:

Problem 1: Take the derivative of the function R(x), set it equal to 0, solve for x. That X gives you the maximum value (or minimum, it depends on the function, but maximum in this case) of R(X) (if the derivative of R(x)=0, that means this is a turn-over point at which R(x) reaches a maxsimum or minimum value)

Problem 2: a) R(x)=p(x)*x=500*x-x^2 (simply the definition of Revenue)
b) it's a graph, I can provide tips if you need help, but I suggest just plotting x and R values on graph paper and connecting them with a line.
c) find the maximum R(X) value from the graph and estimate x
d) use the method from problem 1 to determine where the derivative of R(X) equals 0 and then solve for x to find the maximum revenue

Problem 3: What is the survival function for this problem? Is it in your book? I would suggest integrating the survival function and finding out at what value the integrated survival funtion equals 0.5 and when it equals 1 (or 0), these will be the solutions to the two parts.

I hope this helps! Feel free to ask again or PM me if you want more help, I like calculus and would enjoy trying to help you if you wish.
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Postby Technomancer » Tue Sep 05, 2006 8:02 pm

Puritan wrote:Problem 1: Take the derivative of the function R(x), set it equal to 0, solve for x. That X gives you the maximum value (or minimum, it depends on the function, but maximum in this case) of R(X) (if the derivative of R(x)=0, that means this is a turn-over point at which R(x) reaches a maxsimum or minimum value)


Just to add to this: If you wish to determine whether the point R'(x0)=0 is a maximum or a minimum without looking at the graph, you can do so using the second derivative.

If R''(x0)>0, then R(x0) is a local minimum.
If R''(x0)<0, then R(x0) is a local maximum.

Alternatively, you could also perform the first derivative test, which is similarly straightforward.
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Postby termyt » Fri Sep 08, 2006 5:04 am

I used to love math. It's obviously "used to" because reading this, I realize how much I have forgotten. Was university really that long ago?

*sigh*
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