Primero calculamos el valor de optimo de pedido (Y) para cada caso.
\(Y=\left[\left(\frac{2KD}{h}\right)\right]^{0.5}\)
\(Y_a=\left[\left(\frac{2\left(100\cdot30\right)}{0.05}\right)\right]^{0.5}\)
\(Y_a=346.4101615\)
\(Y_b=\left[\left(\frac{2\left(50\cdot30\right)}{0.05}\right)\right]^{0.5}\)
\(Y_b=244.9489743\)
\(Y_c=\left[\left(\frac{2\left(100\cdot40\right)}{0.01}\right)\right]^{0.5}\)
\(Y_c=894.427191\)
\(Y_d=\left[\left(\frac{2\left(100\cdot20\right)}{0.04}\right)\right]^{0.5}\)
\(Y_d=316.227766\)
Comprobación con uso de la condición:
\(\ \frac{d\ CTU_{\left(Y\right)}}{dy}=-\left(\frac{KD}{y^2}\right)+\left(\frac{h}{2}\right)=0\)
\(\ \frac{d\ CTU_{a\left(Y\right)}}{dy}=-\left(\frac{100\cdot30}{346.4101615^2}\right)+\left(\frac{0.05}{2}\right)\)
\(\frac{d\ CTU_{a\left(Y\right)}}{dy}=-0.025+0.025=0\)
\(\ \frac{d\ CTU_{b\left(Y\right)}}{dy}=-\left(\frac{50\cdot30}{244.9489743^2}\right)+\left(\frac{0.05}{2}\right)\)
\(\frac{d\ CTU_{b\left(Y\right)}}{dy}=-0.025+0.025=0\)
\(\ \frac{d\ CTU_{c\left(Y\right)}}{dy}=-\left(\frac{100\cdot40}{894.427191^2}\right)+\left(\frac{0.01}{2}\right)\)
\(\frac{d\ CTU_{c\left(Y\right)}}{dy}=-0.005+0.005=0\)
\(\ \frac{d\ CTU_{d\left(Y\right)}}{dy}=-\left(\frac{100\cdot20}{316.227766^2}\right)+\left(\frac{0.04}{2}\right)\)
\(\frac{d\ CTU_{d\left(Y\right)}}{dy}=-0.02+0.02=0\)