Consider a set $\mathcal{P}$ defined as $$ \mathcal{P} := \left\{ (x,y) \mid x\in[0,\bar{x}], y\in[0,\bar{y}], xy=0 \right\}.$$ I am trying to prove that $\operatorname{conv}(\mathcal{P}) = \tilde{\mathcal{P}}$ where $\tilde{\mathcal{P}}$ is defined as $$ \tilde{\mathcal{P}} := \left\{ (x,y) \, \bigg| \, x\geq 0, y\geq 0, \frac{x}{\bar{x}} + \frac{y}{\bar{y}} \leq 1 \right\}.$$ I am sure that my hypothesis is correct because simply drawing the two sets on a piece of paper shows that $\tilde{\mathcal{P}}$ should indeed be $\operatorname{conv}(\mathcal{P})$. However, I am struggling to prove it analytically. The approach I'm using is to show that $\mathcal{\tilde{P}} \subseteq \operatorname{conv}(\mathcal{P})$ and $\operatorname{conv}(\mathcal{P}) \subseteq \mathcal{\tilde{P}}$ both hold simultaneously. However, I have not been able to make much headway, and some help would be appreciated.
(edit) Managed to show that $\tilde{\mathcal{P}} \subseteq \operatorname{conv}(\mathcal{P})$ since $\{ (0,0),(0,\bar{x}),(\bar{y},0) \} \in \mathcal{P}$ and every point in $\tilde{\mathcal{P}}$ belongs to $\operatorname{conv} \{ (0,0),(0,\bar{x}),(\bar{y},0) \}$.
You actually already managed to show the harder inclusion. For the opposite inclusion, it suffices to show that your set $\hat{\mathcal P}$ is convex and contains the given set $\mathcal P$.