Take $6x + 9y \le 480$ and solve it for $y$. Once $y$ is written in terms of $x$, think about what happens to the maximum allowed $y$ when $x$ gets larger or smaller.

Since larger $x$ will reduce how big $y$ can be, try using the smallest $x$ that still satisfies $x + y \ge 40$ and $x \ge 0$. Substitute that $x$ into the inequalities to find the allowed range for $y$, then remember that $y$ must be a whole number.

In Desmos, enter the two inequalities:

• 6x + 9y <= 480
• x + y >= 40

Also, because $x$ and $y$ are numbers of parts, add x >= 0 and y >= 0. You should see a shaded feasible region where all these conditions overlap.

Add the boundary line of the time inequality by entering y = (480 - 6x)/9. This line forms the upper edge of the feasible region in terms of $y$; points above it would use more than 480 minutes.

Add the line x = 0 to represent making only type B parts. Tap the intersection point of x = 0 and y = (480 - 6x)/9 to see its $y$-coordinate. That value is the maximum continuous (not necessarily whole-number) $y$ allowed; the greatest whole-number $y$ you can actually produce is the largest integer less than or equal to that $y$-value that still lies in the shaded region.

We are told:

• Each type A part takes 6 minutes, so $6x$ minutes for $x$ parts.
• Each type B part takes 9 minutes, so $9y$ minutes for $y$ parts.
• The machine runs at most 480 minutes, so

• At least 40 total parts must be made, so

Also, $x$ and $y$ represent numbers of parts, so they must be nonnegative whole numbers.

Start with the time inequality and solve for $y$:

This tells us the maximum possible $y$ for any given $x$.

Notice the $-\frac{2}{3}x$ term: as $x$ increases, the right-hand side gets smaller. That means increasing $x$ will decrease the maximum possible $y$.

Since increasing $x$ reduces how large $y$ can be, we want the smallest $x$ that still satisfies all conditions.

The smallest possible $x$ is $0$ (you cannot make a negative number of type A parts).

Check the part-count inequality when $x=0$:

So $x=0$ is allowed as long as $y \ge 40$.

Now apply $x=0$ to the time inequality:

Combining both conditions for $y$ when $x=0$ gives a range:

The machine cannot produce a fraction of a part, so $y$ must be an integer.

From the previous step, $y$ must satisfy $40 \le y \le \frac{160}{3}$, and $\frac{160}{3} \approx 53.3$. The greatest whole number less than or equal to $53.3$ is $53$.

Therefore, the greatest possible value of $y$ is 53.