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?
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
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
To find the number of bars for which the revenue is at a maximum and the unit revenue at that level of production, we need to take a few steps. Here's how we can do it:
Step 1: The given function R(x) represents the unit revenue in dollars for producing x bars. The expression -0.06x² + 10.2x - 50 is a quadratic function in the form of ax² + bx + c.
Step 2: To find the number of bars for which the revenue is at a maximum, we need to determine the vertex of the quadratic function. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a = -0.06 and b = 10.2.
Calculating:
x = -10.2 / (2 * -0.06)
x = -10.2 / -0.12
x = 85
Step 3: Therefore, the revenue is at a maximum when the company produces 85 bars.
Step 4: To find the unit revenue at that level of production, substitute x = 85 into the function R(x).
Calculating:
R(85) = -0.06(85)² + 10.2(85) - 50
R(85) = -0.06(7,225) + 867 - 50
R(85) = -433.5 + 817
R(85) = 383.5
Step 5: Therefore, the unit revenue at the level of production when the company produces 85 bars is $383.5.
In summary:
- The revenue is at a maximum when the company produces 85 bars.
- The unit revenue at that level of production is $383.5.