Answer:
$53
Step-by-step explanation:
[tex]n=\frac{5}{x-6}+5*(100-x)[/tex]
Let 'x' be the selling price. The revenue is given by 'xn' while the production cost is given by '6n' therefore, the profit function is:
[tex]P=xn-6n=(x-6)*(\frac{5}{x-6}+500-5x))\\P=5+500x-3000-5x^2+30x\\P=-5x^2+530x-2950[/tex]
The price 'x' for which the derivate of the profit function is zero is the price that maximizes profit:
[tex]P=-5x^2+530x-2950\\P'=0=-10x+530\\x=\$53[/tex]
The selling price that will maximize profit is $53.
