Calculus-Optimization Problems

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SUMMARY

The discussion focuses on optimizing the number of conversations at a social party modeled by the function N(t) = 30t - t², where t represents time in minutes. The first derivative, N'(t) = 30 - 2t, is used to find the critical point, leading to t = 15 minutes as the time when the maximum number of conversations occurs. Substituting t back into the original function yields a maximum of 225 separate conversations. This analysis confirms the effectiveness of calculus in solving optimization problems.

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Homework Statement


A study has determined that interactions at a large social party 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 occurring. At what time in a party do the most conversations occur? what is the maximum number of interactions?


Homework Equations


The Product Rule
F(x)=f(x)g(x) then F'(x)=f(x)g'(x)+f'(x)g(x)

The Chain Rule for Polynomials
F(x)=(f(x))^n,then F'(x)=nf'(x)f(x)^n-1


The Attempt at a Solution



So i decided to get the 1st derivative of N(t)=30t-t^2
N'(t)=30-2t
so now I'm trying to find t when the function equals 0.
0=30-2t
-30/-2=t
t=15 so this is the time since the party began.

now this is the part I'm not so sure of myself, i add the value of t to the 1st function
N(15)=30(15)-(15)^2
=450-225
=225 so yeah i don't understand the result, what does this one equal to? I'm just confused on how to obtain the results i need here.
 
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225 is N(15). I guess that's the 'maximum number of separate conversations occurring'. I think you've solved the problem.
 
lol, see, that is what happens when you doubt yourself all the time like i do hahaha.
 

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