SOLUTION: The profit of a retail store is y=2500+36x-0.2x^2 . Use the derivative if it would be profitable for the daily advertising budget to be increased if the daily advertising budget is

Question 1056699: The profit of a retail store is y=2500+36x-0.2x^2 . Use the derivative if it would be profitable for the daily advertising budget to be increased if the daily advertising budget is
(a) $60
(b) $300
(c) what is the maximum value for x below which it is profitable to increase the advertising?
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
I'll do the answers backwards,
c)

An increase in profits occurs as long as the derivative is greater than zero.

b) , yes
a) ,no
RELATED QUESTIONS
Let X be the amount (in hundreds of dollars) a company spends on advertising, and let P... (answered by stanbon)

The table shows the daily production level and profit for a business. Use the quadratic... (answered by MathLover1)

A store owner average day sales of $500, realizing a profit of 20% of the selling price.... (answered by Theo)

Use the Quadratic Formula to find the zeros of the equation.... (answered by solver91311)

a retail store with a monthly advertising budget of $3400 decides to set up a media... (answered by josmiceli)

A factory finds that it's daily profit, y dollars, is related to x, the number of items... (answered by mananth)

Daily profits at a retail store have an average of $1000. If at least 88.9% of values are (answered by stanbon)

a store increases sales $5000 for every $1500 it spends on advertising how much will... (answered by Cromlix,josmiceli,Edwin McCravy)

Hello! I'd greatly appreciate it if anyone can help me on this deceptively hard (for me,... (answered by vleith,Greenfinch)