Let `d` be diagonal and `a` and `b` sides of TV. By Pythagorean theorem

`d=sqrt(a^2+b^2)`

Also since `a:b=16:9` it follows `a=16/9b`and `b=9/16a` hence

`d=sqrt((16/9b)^2+b^2)=sqrt(337)/9b` **(1)**

So if we put `b=18.5` (we subtract 1in from both sides for the frame) we will get upper bound on...

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Let `d` be diagonal and `a` and `b` sides of TV. By Pythagorean theorem

`d=sqrt(a^2+b^2)`

Also since `a:b=16:9` it follows `a=16/9b`and `b=9/16a` hence

`d=sqrt((16/9b)^2+b^2)=sqrt(337)/9b` **(1)**

So if we put `b=18.5` (we subtract 1in from both sides for the frame) we will get upper bound on TV diagonal

`d leq sqrt(337)/9 cdot 18.5 approx 37.73` **(2)**

Since `b=9/16a` we can write (1) as

`d=sqrt(337)/16a`

Now to get upper bound on `d` we put `a=40`

`d leq sqrt(337)/16 cdot40 approx 45.89` **(3)**

Now taking both (2) and (3) into account we see that** maximum size of TV diagonal is 37.73 in.**