A company producing steel construction bars uses the function R(x) = -0.06x²+10.2x -50 to model the unit revenue in dollars for producing x bars. For what number of bars is the revenue at a maximum? What is the unit revenue at that level of production?

Question

A company producing steel construction bars uses the function R(x) = -0.06x²+10.2x -50 to model the unit revenue in dollars for producing x bars.  For what number of bars is the revenue at a maximum?  What is the unit revenue at that level of production?

Damon · Answer

if you do calculus
dR/dt = -0.12 x + 10.2
= 0 for max or min
x = 10.2 /0.12 = 85 bars
unit rev for 85 bars = R(85) / 85

If you do not know calculus, complete the square for x and R to find the vertex of
-0.06x²+10.2x -50 = R

Rebecca · Answer

My course is College Algebra concepts. We are dealing with quadratic equations. I got the first part, but I am having a hard time with the second part of the question.
x=-b/2a
x=-10.2/2(-0.06)
x=-10.2/(-0.12)=-(-85)=85
x=85 bars
So I would simply take R(85) / 85 giving me 1 or replace x with 85?
R(85) = -0.06(85)^2+10.2(85) -50
R(85)= -0.06(7225)+867 -50
R(85)= -433.5+867 -50
R(85)=383.5
Sorry for all the quesions.
Rebecca