Question 464081
C(1) = 100(100 + 9 + 144) = 25,300 $ total cost of storage and transportation.

{{{C(k)=100(100+9k+144/k)}}} ==> {{{dC(k)/dk = 100(9-144/k^2)}}}.

Setting the derivative equal to 0,
 {{{ 100(9-144/k^2) = 0}}} ==> {{{9 - 144/k^2 = 0}}} ==> {{{9 = 144/k^2}}}

==> {{{k^2 = 144/9}}}  ==> k = 4.

Now {{{ d^2C(k)/dk^2 = 100(288/k^3) > 0}}} when k = 4.

==> Absolute minimum at k = 4.  

Minimum value is C(4) = 100(100 + 9*4 + 144/4) = 17,200 $.