Step 1
Let the dimensions be $ x $ cm length and $ y$ cm breadth.
Area of poster = $ A = xy$
Step 2
Given : (x-8)(y-4)=50$
or $ y = \large\frac{50}{x-8}+4$
$ \Rightarrow A = \bigg[ \large\frac{50}{x-8}+4 \bigg]^x$
Step 3
$ \large\frac{dA}{dx}=4-\large\frac{400}{(x-8)^2}=0$
Ans $ \Rightarrow x = 18, \: \: y = 9$
Step 4
$ \large\frac{d^2A}{dx^2}=$ positive $ \Rightarrow A$ is minimum.