Graph the revenue function $R(p) = -\frac{1}{2}(p-4)^2(p-10)+32$ and look for the highest point on the curve between $p = 4$ and $p = 10$.

For a function on a closed interval like $[4, 10]$, the maximum occurs either at an endpoint or at a turning point. Check $R(4)$, $R(10)$, and any peaks you see on the graph.

In Desmos, type the function using $x$ as the variable:

R(x) = -1/2 * (x - 4)^2 * (x - 10) + 32

This will graph the revenue versus the price.

Adjust the x-axis so you can clearly see from $x = 4$ to $x = 10$. Adjust the y-axis so that all relevant revenue values are visible (for example, 0 to 60).

Click on the graph near its highest point between $x = 4$ and $x = 10$. Desmos will show the coordinates. The x-coordinate of this highest point is the price $p$ that maximizes the daily revenue.

The function

gives the daily revenue (in thousands of dollars) when the price is $p$ dollars, with $4 \le p \le 10$.

We want the price $p$ in this interval that makes $R(p)$ as large as possible.

First, evaluate the revenue at the endpoints of the interval:

• At $p = 4$: $(p-4)^2 = 0$, so $R(4) = 0 + 32 = 32$.
• At $p = 10$: $(p-10) = 0$, so $R(10) = 0 + 32 = 32$.

Both endpoints give the same revenue of 32 thousand dollars.

Graph $R(p) = -\frac{1}{2}(p-4)^2(p-10)+32$ in Desmos. Looking at the graph between $p = 4$ and $p = 10$, you can see the curve rises above 32 and has a peak somewhere in the middle.

Use Desmos's maximum feature or click on the highest point of the curve. You'll find the maximum occurs at $p = 8$.

At $p = 8$:

• $(8-4)^2 = 16$
• $(8-10) = -2$
• $R(8) = -\frac{1}{2}(16)(-2) + 32 = 16 + 32 = 48$

Since $R(8) = 48 > R(4) = R(10) = 32$, the maximum revenue of 48 thousand dollars occurs when $p = 8$.

The daily revenue is maximized when the subscription price is 8 dollars.