Is it possible to cover an $11 \times 12$ rectangle with $19$ rectangles of $1 \times 6$ or $1 \times 7$?

$$\matrix{\color{red}1,0,0,0,0,0,0,\color{red}1,0,0,0,0\\0,\color{red}1,0,0,0,0,0,0,\color{red}1,0,0,0\\0,0,\color{red}1,0,0,0,0,0,0,\color{red}1,0,0\\0,0,0,\color{red}1,0,0,0,0,0,0,\color{red}1,0\\0,0,0,0,\color{red}1,0,0,0,0,0,0,\color{red}1\\0,0,0,0,0,\color{red}1,0,0,0,0,0,0\\0,0,0,0,0,0,\color{red}1,0,0,0,0,0\\\color{red}1,0,0,0,0,0,0,\color{red}1,0,0,0,0\\0,\color{red}1,0,0,0,0,0,0,\color{red}1,0,0,0\\0,0,\color{red}1,0,0,0,0,0,0,\color{red}1,0,0\\0,0,0,\color{red}1,0,0,0,0,0,0,\color{red}1,0} $$

We have $20$ red cells and $19$ rectangles. Each rectangle can cover at most $1$ red cell. So..

Same as @Aqua's answer but I made a pretty image:

No two rectangles can cover more than one red square, so you need at least $20$ rectangles to cover all red squares.