People arrive at a college admissions office at rate 1 per minute, and the arrival is a Poisson process. When k people have arrive, a tour starts. Student tour guides are paid $20 for each tour they conduct. The college estimates that it loses ten cents in good will for each minute a person waits. What tour group (not counting the guide) minimizes the average cost per person?

K=20

Step-by-step explanation:

There seem to be no randomness in the question.

At 1 per minutes the arrival rate is fixed.

Then compute the average cost for each person to give a four, adding the cost of guide and time waiting cost..

Therefore, K is the number of people hoping will show up.

Number of per minute waiting

= 1/2(K-1)K.

Tour cost 20+1/20(K-1).

Cost per guest= 20/k +1/20(K-1)

If the derivative is set to Zero

K=20

step-by-step explanation:

when the line is solid:

when the line is dashed:

this > inequality sign means:

this < inequality sign means:

this > inequality sign means:

this < inequality sign means:

about the problem:

srry i'm not done ! i'll

my guess would be b and d, but i didn’t actually get out the pen and paper. lol