# Determine the number of units x that produce a maximum revenue, in dollars, for the given revenue...

## Question:

Determine the number of units {eq}x {/eq} that produce a maximum revenue, in dollars, for the given revenue function. Also, determine the maximum revenue.

$$R(x) = 810-0.6x^2$$

## Extrema:

At the end of the day the math is just math, it has no face; that's what makes it so useful. Whenever we want to maximize or minimize a function (no matter what it describes), we can always use differentiation. Since the derivative is the slope of the tangent line to a curve at any point, we can find extrema by computing the derivative, then seeing where it equals 0 or is undefined.

We want to maximize a function; we will do so using differentiation. The derivative of the revenue (a.k.a. the marginal revenue) is

{eq}\begin{align*} R' (x) &= \frac{d}{dx} \left( 810-0.6x^2 \right) \\ &= 1.2 x \end{align*} {/eq}

And we see right away that this is not a realistic equation at all! Clearly this function equals 0 when {eq}x = 0 {/eq}, which means we maximize revenue by producing nothing, and that the maximum revenue is

{eq}\begin{align*} R(0) &= \\$810 \end{align*} {/eq}

Again it needs to be stressed how unrealistic this is; most revenue functions have a linear term along with the negative quadratic term. It is very likely that this revenue function should have been {eq}R (x) = 810x - 0.6x^2 {/eq}; at least then we would have had a realistic result.