You want the annual permit to cost less than the pay-as-you-go plan. That means you should set up an inequality where $480$ is on one side and the pay-as-you-go expression is on the other. Start with the case $d\le 100$.

When you solve the inequality for the first 100 days, you will get a condition like "$d$ is greater than some number." Remember that $d$ must be a whole number of days. What is the smallest whole number of days that satisfies this?

After you handle $d\le 100$, also consider $d>100$ with the cheaper $2$-dollar rate. Does this give you an even smaller number of days, or are those days all larger than what you already found?

In Desmos, enter the pay-as-you-go cost for the first 100 days as y = 6x {0 < x <= 100}. Then enter the cost after 100 days as y = 2x + 400 {x > 100}. These two pieces together represent the total pay-as-you-go cost.

Enter the annual permit cost as a horizontal line: y = 480. This line shows the flat cost of the permit for any number of days $x$.

Look for the $x$-value where the pay-as-you-go graph intersects the horizontal line $y = 480$ while $x \le 100$. That $x$-value is where the two plans cost the same. To answer the question, think about the smallest whole number of days greater than this $x$-value, because the permit must be strictly cheaper, not just equal in cost.

Let $d$ be the number of days the student parks on campus during the academic year.

Annual permit:

• Costs a flat $480$, no matter how many days are used.

Pay-as-you-go:

• For the first 100 days, the cost is $6$ dollars per day, so for $d\le 100$ the cost is $6d$.
• For more than 100 days, the first 100 days cost $6\cdot 100=600$ dollars, and each additional day costs $2$ dollars.
  • For $d>100$, extra days are $d-100$, so cost is $600+2(d-100)=2d+400$.

We want the annual permit to cost less than the pay-as-you-go plan.

For $d\le 100$:

• Pay-as-you-go cost is $6d$.
• Annual permit is $480$.

Set up the inequality:

• $480<6d$ (permit cost less than pay-as-you-go).

Solve for $d$:

• Divide both sides by $6$: $80<d$.

So, in the range $d\le 100$, the annual permit is cheaper whenever $d$ is greater than 80.

Now consider $d>100$.

For $d>100$:

• Pay-as-you-go cost is $2d+400$.
• Annual permit is still $480$.

Set up the inequality again:

• $480<2d+400$.

Solve for $d$:

• Subtract $400$ from both sides: $80<2d$.
• Divide by $2$: $40<d$.

But in this case we already know $d>100$, which automatically satisfies $40<d$. So for every day after 100, the annual permit is also cheaper.

From the $d\le 100$ case, the permit is cheaper when $d>80$.

• The smallest whole number of days greater than $80$ is $81$.

Check around this point:

• At $d=80$: pay-as-you-go is $6\cdot 80=480$, which is equal to the permit, not more.
• At $d=81$: pay-as-you-go is $6\cdot 81=486$, which is more than $480$.

Therefore, the minimum number of days a student must park so that the annual permit costs less than pay-as-you-go is 81.