Answer:
[tex]x \in (40,110)[/tex]
Step-by-step explanation:
We are given the following in the question:
[tex]p(x)=-x ^2 + 150 x - 4400[/tex]
where p gives the profit.
We want to keep the profit positive, thus
[tex]p(x)=-x ^2 + 150 x - 4400 > 0\\-x ^2 + 150 x - 4400 > 0\\x^2-150x+4400<0\\x^2-110x-40x+4400<0\\(x-110)(x-40)<0\\x<110,x<40\\x \in (40,110)[/tex]
Thus, x should lie between 40 and 110 so that the profit is always positive.
