MATH SOLVE

7 months ago

Q:
# The profits of mr cash’s company is represented by the equation p(t)=-3t^2+18t-4, where p(t) is the amount of profit in hundreds of thousands of dollars and t is the number of years of operation. he realizes his company is on the down turn and wishes to sell before he ends up in debt. In what year of operation does Mr. Cash’s business show maximum profit? What is the maximum profit? What time will it be too late to sell business?

Accepted Solution

A:

Answer:$2300000- max profit5.77 years into operation - zero profit timeStep-by-step explanation:Given function:p(t)= -3t^2 + 18t - 4It is a quadratic function with general form of y = ax^2 + bx + cIt opens down if a < 0 and gets maximum value at vertex which is determined at x = -b/2aFor the given function vertex is:t = -18/-3*2 = 3Maximum value of p is:p(3) = -3*3^2 + 18*3 - 4 = -27 + 54 - 4 = 23 ($2300000)The time when profit is zero:-3t^2 + 18t - 4 = 03t^2 - 18t + 4 = 0t = (18 ± √18² - 4*3*4)/2*3 = (18 ± √276)/6 = (18 ± 16.61)/6t = (18 - 16.61)/6 = 0.23 this is the time when company has just started to make profit, it is not applicable as is the past timet = (18 + 16.61)/6 = 5.77 years into operation, after this time there will be no profit