The values of price p for the product to remain profitable must be less than 1 and 25.
The value of P that will maximize the profit is 13.
Maximum function
For the given product to remain profitable, the revenue must continue to be greater than the cost that is R(p) > C(p)
Given the following parameters:
- [tex]R(p) = - p^2+ 30p\\C(p)=4p+25[/tex]
If R(p) > C(p), then;
[tex]- p^2+ 30p>4p+25\\p^2-26p+25<0[/tex]
Factorize to get the value of p
[tex]p^2-p-25p+25<0\\p(p-1)-25(p-1)<0\\(p-1)(p-25)<0\\p<1 \ and \ p<25[/tex]
Hence the values of price p for the product to remain profitable must be less than 1 and 25
To maxmimize the profit, dP/dp = 0
Profit = Revenue - Cost
Profit = [tex]- p^2+ 30p - 4p - 25[/tex]
Profit = [tex]-p^2+26p-25[/tex]
Differentiate the function and equate to zero;
[tex]dP/dp = -2p+26\\-2p + 26 = 0\\-2p = -26\\p=13[/tex]
Hence the value of P that will maximize the profit is 13
Learn more on cost and revenue functions here: https://brainly.com/question/25638609
