Answer:
the EOQ cost function [tex]\mathbf{g(Q)= \dfrac{h}{2 \lambda Q}( Q- \sqrt{\dfrac{2 \lambda K }{2}})^2 + \sqrt{\dfrac{2 h K}{\lambda}}}[/tex]
Step-by-step explanation:
Given that:
[tex]g(Q) = \dfrac{h}{2 \lambda Q} \begin {pmatrix} Q- \sqrt{\dfrac{2 K \lambda}{h}} \end {pmatrix}^2 + \sqrt{\dfrac{2 Kh}{\lambda}}[/tex]
Total cost = Purchase Cost + Ordering Cost + Holding Cost
i.e
T = [tex]P \lambda + h \dfrac{Q}{2}+\dfrac{\lambda K}{Q}[/tex]
By differentiating with respect to Q and equating to zero, we have:
[tex]0 = 0 +\dfrac{h}{2}- \dfrac{\lambda K}{Q^2}[/tex]
[tex]Q* = \sqrt{\dfrac{2 k \lambda}{h}}[/tex]
Now;
[tex]T = P \lambda + \dfrac{h}{2Q}(Q-Q^*)^2 + hQ^*[/tex]
[tex]T = \dfrac{h}{2Q}(Q- \sqrt{\dfrac{2 \lambda K}{h}})^2 + \sqrt{2 hK \lambda } + P \lambda[/tex]
[tex](\dfrac{T - P \lambda}{\lambda }) = \dfrac{h}{2 \lambda Q}( Q- \sqrt{\dfrac{2 \lambda K }{2}})^2 + \sqrt{\dfrac{2 h K}{\lambda}}[/tex]
Therefore:
the EOQ cost function [tex]\mathbf{g(Q)= \dfrac{h}{2 \lambda Q}( Q- \sqrt{\dfrac{2 \lambda K }{2}})^2 + \sqrt{\dfrac{2 h K}{\lambda}}}[/tex]
