For a smooth function like this one, the maximum on $[4,10]$ will occur either at an endpoint ($p=4$ or $p=10$) or at an interior point where the graph changes from going up to going down (where the slope is zero). Identify these candidate points.

One algebraic way to find interior candidates is to take the derivative of $R(p)$, set it equal to 0, and solve the resulting equation for $p$. Then compare the revenue at those $p$-values with the revenue at $p=4$ and $p=10$.

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 (in thousands of dollars) versus the price $x$.

Adjust the x-axis so you can clearly see from $x=4$ to $x=10$ (for example, set the x-axis from 0 to 12). Adjust the y-axis so that all relevant revenue values are visible (for example, 0 to 60). You should now see the curve of $R(x)$ over the interval $[4,10]$.

Use Desmos’s maximum feature: either click on the graph near its highest point between $x=4$ and $x=10$ and choose “maximum,” or type maximum(R, 4, 10) in a new line. Read off the x-coordinate of this highest point on the graph; that x-value 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 (maximizes the revenue). For a smooth function on a closed interval, the maximum can occur:

• at an endpoint ($p=4$ or $p=10$), or
• at an interior point where the graph changes from increasing to decreasing (where the slope is $0$).

It can be helpful to expand the expression so we can work with standard polynomial tools.

First expand $(p-4)^2$:

Now multiply by $(p-10)$:

Multiply by $-\tfrac12$ and then add $32$:

So an equivalent form is

The revenue will be maximized at a point where the curve stops going up and starts going down, which happens where the slope (derivative) is $0$.

Differentiate $R(p)$ term by term:

• The derivative of $-\tfrac12 p^3$ is $-\tfrac32 p^2$.
• The derivative of $9p^2$ is $18p$.
• The derivative of $-48p$ is $-48$.
• The derivative of $112$ is $0$.

So the slope function is

Set this equal to zero to find critical points:

Multiply both sides by $-2$ to clear the fraction:

Divide by $3$:

The quadratic equation from the previous step,

has two solutions. These two $p$-values are critical points where the slope is zero.

Any maximum of $R(p)$ on $[4,10]$ must occur at one of these two critical points or at an endpoint of the interval ($p=4$ or $p=10$). So the candidate prices to check are:

• the two solutions of $p^2 - 12p + 32 = 0$, and
• $p=4$ and $p=10$.

First, solve the quadratic

We look for two numbers that multiply to $32$ and add to $-12$: they are $-4$ and $-8$. So

which gives critical points $p=4$ and $p=8$.

Now evaluate the revenue at all candidates (remember $R(p)=-\tfrac12(p-4)^2(p-10)+32$):

• At $p=4$:
  • $(p-4)^2 = 0$, so $R(4) = 32$.
• At $p=8$:
  • $(8-4)^2 = 16$, $(8-10) = -2$,
  • $-\tfrac12 \cdot 16 \cdot (-2) = 16$, so $R(8) = 16 + 32 = 48$.
• At $p=10$:
  • $(p-10) = 0$, so $R(10) = 32$.

The largest revenue value is $48$ (thousand dollars), which occurs when $p=8$. Therefore, the daily revenue is maximized when the subscription price is $\boxed{8}$ dollars.