For the plan without membership, split the cost into the first 100 posters and any posters after 100. What does the cost look like when $p > 100$?

You want the membership plan to be cheaper. Using the expression for the regular plan when $p > 100$, write an inequality of the form (membership cost) < (regular cost) and solve it for $p$.

After solving the inequality, the solution will not be a whole number. Think about which integer values of $p$ satisfy the inequality and choose the smallest such integer.

In Desmos, type y1 = 210 + 0.85x for the membership plan. For the regular plan when more than 100 posters are ordered, type y2 = x + 65.

Use Desmos to find where the two lines intersect by clicking on their intersection; note the $x$-value of this point. This $x$-value is where the two plans cost the same, and membership becomes cheaper for $x$-values greater than this.

Since the number of posters must be an integer and the membership must cost less, take the smallest whole-number $x$-value that is greater than the intersection $x$-value shown in Desmos.

Let $p$ be the number of posters ordered.

• Membership plan: no setup fee, plus a $210$ membership fee, plus $0.85$ per poster:
  • Cost with membership: $210 + 0.85p$.
• Without membership, the print shop charges:
  • $40$ setup fee,
  • $1.25$ each for the first $100$ posters,
  • $1.00$ for each poster after the first $100$.

For the plan without membership:

• If $p \le 100$:
  • All posters are in the “first 100,” so cost is $40 + 1.25p$.
• If $p > 100$:

  • First 100 posters cost $1.25$ each: $100 \times 1.25 = 125$.

  • Remaining $p - 100$ posters cost $1.00$ each: $1.00(p - 100)$.

  • Total cost:

So for $p > 100$, no-membership cost is $p + 65$.

First, check whether membership can be cheaper when $p \le 100$ by comparing

• Membership: $210 + 0.85p$
• No membership: $40 + 1.25p$.

Set up the inequality for $p \le 100$:

Solve it:

But this contradicts $p \le 100$. So membership can only be cheaper when $p > 100$, where the no-membership cost is $p + 65$.

Now compare costs for $p > 100$:

• Membership: $210 + 0.85p$
• No membership: $p + 65$

We want membership to cost less:

Solve step by step:

Compute $\frac{145}{0.15}$:

So $p$ must be greater than about $966.67$.

The number of posters $p$ must be a whole number, and membership must be strictly cheaper, so we need the smallest integer greater than $966.67$.

That integer is $p = 967$.

So, the customer must order 967 posters for the membership plan to cost less than ordering without the plan.