To get minimized number of x steel bars, we differentiate the equation and then equate to zero:
d/dx (C(x) = 0.02x² – 3.4x + 150)
C'(x) = 0.02(2)x – 3.4 = 0
Solving for x
0.04x – 3.4 = 0
x = 85 steel bars
For minimum cost,
C(x = 85) = 0.02(85)² – 3.4(85) + 150 = 5.5 dollars
