We can express the defective panels (D) multypling the production (P) by the defective rate (f):
![D=f\cdot P](https://img.qammunity.org/2023/formulas/mathematics/college/jptq5es8a4sqdmtqcgme7gbaawgeuumkbo.png)
This can be applied to the total production as well as the production in each plant:
![\begin{gathered} D_a=f_a\cdot P_a \\ D_b=f_b\cdot P_b \\ D=f\cdot P \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/v29qmtvu7awegszdt94gikhy9hl0joz74p.png)
Also, we know that the total production P is equal to the sum of the production of both plants:
![P=P_a+P_b](https://img.qammunity.org/2023/formulas/mathematics/college/h5ausfm1ipebcbhuhihswlt079pukmm4ji.png)
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:
![\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=(f-f_a)/(f_b-f)\cdot P_a \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/gviqid5eovugmhr74b786t1y240ksz3ljk.png)
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:
![\begin{gathered} P_b=(f-f_a)/(f_b-f)\cdot P_a \\ P_b=(0.08-0.03)/(0.10-0.08)\cdot8000 \\ P_b=(0.05)/(0.02)\cdot8000 \\ P_b=2.5\cdot8000 \\ P_b=20000 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/a7tn1klj14awdtxnlpq5uwv7ihwhqvvp2o.png)
Answer: The production of plant B is 20000 panels.