Respuesta :
We can express the defective panels (D) multypling the production (P) by the defective rate (f):
[tex]D=f\cdot P[/tex]This can be applied to the total production as well as the production in each plant:
[tex]\begin{gathered} D_a=f_a\cdot P_a \\ D_b=f_b\cdot P_b \\ D=f\cdot P \end{gathered}[/tex]Also, we know that the total production P is equal to the sum of the production of both plants:
[tex]P=P_a+P_b[/tex]We can find the production of Plant B by stating that the sum of the defective panels of each plant is equal to the total defective panels:
[tex]\begin{gathered} D_a+D_b=D \\ f_aP_a+f_bP_b=fP \\ f_aP_a+f_bP_b=f(P_a+P_b) \\ f_aP_a+f_bP_b=fP_a+fP_b \\ f_bP_b-fP_b=fP_a-f_aP_a \\ (f_b-f)\cdot P_b=(f-f_a)P_a \\ P_b=\frac{f-f_a}{f_b-f}\cdot P_a \end{gathered}[/tex]We know have an expression for the production of Plant B in function of the defective rates and the production of Plant A.
We replace with the values and solve for Pb:
[tex]\begin{gathered} P_b=\frac{f-f_a}{f_b-f}\cdot P_a \\ P_b=\frac{0.08-0.03}{0.10-0.08}\cdot8000 \\ P_b=\frac{0.05}{0.02}\cdot8000 \\ P_b=2.5\cdot8000 \\ P_b=20000 \end{gathered}[/tex]Answer: The production of plant B is 20000 panels.
