How do i minimize this function?

Question

Cost = 21L + (2600/L)

Answer 1

you can graph it and find the minimum or set the derivative equal to zero.

the derivative is:
C'(L)=21- (2600/L^2)

set it equal to zero and you find the critical point
21- (2600/L^2)=0

L= (10√546)/(21)

so in C(L), the function has a minimum when L=(10√546)/(21) and therefore the cost is minimized

Answer 2

You find the derivative and set it equal to zero.
C(L) = 21L + 2600/L
C(L) = 21L + 2600L^(-1)
So ... C'(L) = 21 - 2600L^(-2)
So ... 0 = 21 - 2600/(L^2)
2600/(L^2) = 21
Cross multiply. You get 21L^2 = 2600
L^2 = 123.8095
L = 11.13 (appx.)