A baker has 60 kg of flour and 40 kg of sugar…

Step 1. The model. Let $x$ be the trays of cookies and $y$ the trays of cakes. Maximise $Z = 90x + 120y$ rupees subject to $3x + 2y \le 60$, $x + 2y \le 40$, $x \ge 0$ and $y \ge 0$.

Step 2. The corners. Subtracting the second equation from the first gives $2x = 20$, so $x = 10$, and then $y = 15$. The other corners are $(0, 0)$, $(20, 0)$ and $(0, 20)$.

Step 3. The profit. At $(0, 0)$, $Z = 0$. At $(20, 0)$, $Z = 1800$. At $(10, 15)$, $Z = 900 + 1800 = 2700$. At $(0, 20)$, $Z = 2400$.

Step 4. Say it. Bake $10$ trays of cookies and $15$ trays of cakes. The profit is $2700$ rupees.