A study has determined that the interaction 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? optimization problem, calculus.

Question

anonymous · Answer

no, not calculus

khally92 · Answer

optimization problem

anonymous · Answer

$$N(t)=30t-t^2$$ is a parabola that opens down
it will have max at the second coordinate of the vertex.
first coordinate of the vertex of a parabola is $-\frac{b}{2a}$ in your case $-\frac{30}{2\times -1}=15$

anonymous · Answer

so max occurs when $t=15$ and the maximum number of interactions is 
$$N(15)=30\times 15-15^2$$

anonymous · Answer

you can use calc if you like, seems like overkill
you would write 
$$N(t)=30t-t^2$$
$$N'(t)=30-2t$$ set is equal to zero to get the critical point write 
$$30-2t=0$$$$30=2t$$$$t=15$$

anonymous · Answer

in fact calc is a quick way to show that the first coordinate of the vertex of $y=ax^2+bx+c$ is $-\frac{b}{2a}$ 
$$y'=2ax+b$$ set equal zero and solve, get the critical point at $x=-\frac{b}{2a}$

khally92 · Answer

so t is 15 and N IS 225.