1. **Stating the problem:** We are given two inequalities involving $x$ and $y$: $$2x + 3y < 60$$ $$4x + 3y < 96$$ We want to understand why at $x=0$, these inequalities imply $y < 20$ and $y < 32$, and why $y=20$ is chosen. 2. **Substitute $x=0$ into the inequalities:** For the first inequality: $$2(0) + 3y < 60 \implies 3y < 60 \implies y < \frac{60}{3} = 20$$ For the second inequality: $$4(0) + 3y < 96 \implies 3y < 96 \implies y < \frac{96}{3} = 32$$ 3. **Interpretation:** At $x=0$, the first inequality restricts $y$ to be less than 20, and the second restricts $y$ to be less than 32. 4. **Choosing $y=20$:** Since both inequalities must be satisfied simultaneously, $y$ must be less than both 20 and 32. The more restrictive condition is $y < 20$. Therefore, the maximum value $y$ can approach (but not reach) at $x=0$ while satisfying both inequalities is just under 20. Hence, $y=20$ is the boundary value from the first inequality at $x=0$. This explains why $y=20$ is the critical value at $x=0$ for these inequalities.