# optimization of a social gathering using N(t)=30t-t^2

• Mar 26th 2010, 03:04 PM
surffan
optimization of a social gathering using N(t)=30t-t^2
The question is
the interactions at a social gathering follow the mathematical progression N(t)=30t-t^2, where t is the time in minutes since the party began, and N is the number of separate conversations occuring. what is the maximum number of interactions? thats all the information i was given. I'm not sure what to do with this problem
• Mar 26th 2010, 03:32 PM
craig
Quote:

Originally Posted by surffan
The question is
the interactions at a social gathering follow the mathematical progression N(t)=30t-t^2, where t is the time in minutes since the party began, and N is the number of separate conversations occuring. what is the maximum number of interactions? thats all the information i was given. I'm not sure what to do with this problem

To find the maximum or minimum value of your function you need to differentiate it and set it equal to zero.

In the case above we differentiate wrt t.

\$\displaystyle N(t) = 30t - t^2\$

\$\displaystyle N'(t) = 30 - 2t = 0\$

So \$\displaystyle t = 15\$

You then substitute this value of t back into your original equation to find the maximum value.
• Mar 26th 2010, 03:34 PM
craig
Sorry I should probably mention about minimum values.

If you differentiate your original function, set equal to zero and solve, you may get 1, 2 or inifite solutions, depending on the type of your equation.

To check whether it is a maximum or minimum value you that you have found, you take the second derivative and then substitute in your value(s) again.

If the function is less than 0, then it is a maximum, if it is greater than zero then it is a minimum.

Hope that wasn't too confusing ;)
• Mar 26th 2010, 03:42 PM
surffan
thank you so much, i dont know why i found it so difficult, i kept forgetting to carry the 2 when a function is squared to calculate the derivative. I have another question that is similar except my derivative doesnt leave me with much. its to maximize revenue and the function is 1200(1.50-x)
• Mar 26th 2010, 03:47 PM
craig
Quote:

Originally Posted by surffan
thank you so much, i dont know why i found it so difficult, i kept forgetting to carry the 2 when a function is squared to calculate the derivative. I have another question that is similar except my derivative doesnt leave me with much. its to maximize revenue and the function is 1200(1.50-x)

If you have a new question please in future start a new thread.

However as this is extremely similar to your original I'll just answer it here.

As you will have noticed when you differentiate it, the variable (x) disappears, in this case, there is no maximum as the function is unbounded, so the answer is infinity or undefined.
• Mar 26th 2010, 03:48 PM
surffan
for the function 1200(1.50-x)
this is what i got so far, i dont know if its right
(1800-1200x)x
1800x-1200x^2
f'(x) = 1800-2400x
x=0.75
• Mar 26th 2010, 03:49 PM
craig
Quote:

Originally Posted by surffan
for the function 1200(1.50-x)
this is what i got so far, i dont know if its right
(1800-1200x)x
1800x-1200x^2
f'(x) = 1800-2400x
x=0.75

Where has this extra \$\displaystyle x\$ magically appeared from?
• Mar 26th 2010, 03:52 PM
surffan
I used revenue =price x quantity
the equation represents quantity and the (x) represents price, so i multiplied the two to get an equation for revenue and took the derivative to solve for max and min, and i got 0.75
• Mar 26th 2010, 03:54 PM
craig
In that case then you're correct.