we have
the revenue is equal to
R(X)=[1,500-3x]x-[66,500+20x]
R(x)=1,500x-3x^2-66,500-20x
R(X)=-3x^2+1,480x-66,500
using a graphing tool
the vertex is the point (246.67,116,033.33)
that means
the maximum revenue is For x=246.67
For break even
1,500x-3x^2=66,500+20x
the solutions are
x=50 and x=443
but x=443 is not in the domain
so
break even is for x=50
part e
For x=35
substitute in the revenue equation
R(X)=-3x^2+1,480x-66,500
R(35)=-3(35)^2+1,480(35)-66,500
R(35)=-18,375 -----> is negative, because is a loss
part f
For a profit of 100,000
we have
R(x)=100,000
substitute
100,000=-3x^2+1,480x-66,500
-3x^2+1480x-166500=0
the solution is For x=94.43