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Bounding the area of the image of a set by product of maximum of lengths

Let $F:[0,1]\times[0,1]\to \mathbb {R}^2$ be a smooth function. Given $x\in [0,1]$, let $\ell_x:=\{x\}\times [0,1]$, and given $y\in [0,1]$, let $\ell_y:=[0,1]\times \{y\}$.

My question feels intuitive, by I have failed for a long time to find a rigorous argument. The most naive version of it is the follows:

Is $\mathrm{Area}(\mathrm{Im}(F))\leq \max_{x\in[0,1]}\mathrm{Length}(F[\ell_x])\cdot \max_{y\in[0,1]}\mathrm{Length}(F[\ell_y]) $?

Any drawing I make makes me feel like this should be true, but I cannot prove it. Moreover, it is of interest to me to understand what happens if the image of $F$ is in fact a manifold of bounded curvature? And what if the domain of $F$ is a set which is not a square? E.g $\mathrm{Area}(F[A])\leq \max_{x\in[0,1]}\mathrm{Length}(F[\ell_x\cap A])\cdot \max_{y\in[0,1]}\mathrm{Length}(F[\ell_y\cap A]) $?

Any help, guidance, or reference would be appreciated!