To solve this, we will set up a proportion based on the given rate:

$\frac{50 \, \text{oz}}{4000 \, \text{ft}^2} = \frac{x \, \text{oz}}{9600 \, \text{ft}^2}$

Now, cross-multiply and solve for $x$ (the amount of seed needed for 9600 square feet):

$50 \times 9600 = 4000 \times x$

$480000 = 4000x$

Now, divide both sides by 4000:

$x = \frac{480000}{4000} = 120 \, \text{oz}$

So, you would need 120 ounces of grass seed for 9600 square feet of lawn.

Would you like more details, or do you have any questions?

Here are 5 related questions to explore further:

1. How much seed is needed for a 2000 square foot lawn?
2. How much lawn could you seed with 75 ounces of grass seed?
3. If you reduce the seed amount by 10%, how much would be needed for 9600 square feet?
4. How would the rate change if the lawn condition requires 25% more seed?
5. What if the lawn area is 12000 square feet? How much seed would be required?

Tip: When working with proportions, ensure that the units match on both sides of the equation for consistency.