Apply Mathematics (1)(a. A company’s weekly revenue R is given by the formula R = - p^2+ 30p, where p is the price of the company’s product. The company is considering hiring a distributor, which will cost the company 4p + 25 per week.

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 functionFor 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 25To maxmimize the profit, dP/dp = 0Profit = Revenue - CostProfit = [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 13Learn more on cost and revenue functions here: