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

## 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.

$$R(x) = 296-0.2x^2$$

## Extreme values and Extremity:

Suppose we consider {eq}y=g(x) {/eq} as a function of one variable. Then {eq}x=p {/eq} is called a Critical point of the function if {eq}{\left[ {\frac{{dy}}{{dx}}} \right]_{x = p}}=0 {/eq} and that critical point is called as a point of Minima if {eq}{\left[ {\frac{{{d^2}y}}{{d{x^2}}}} \right]_{x = p}} > 0 {/eq}

Again critical point is called as a point of Maxima if {eq}{\left[ {\frac{{{d^2}y}}{{d{x^2}}}} \right]_{x = p}} < 0 {/eq}.

Here the given function is {eq}R(x) = 296-0.2x^2 {/eq}

Hence,

{eq}\eqalign{ R'\left( x \right) = 0 & \Rightarrow \frac{d}{{dx}}\left[ {296 - 0.2{x^2}} \right] = 0 \cr & \Rightarrow \left[ { - 0.4x} \right] = 0 \cr & \Rightarrow x = 0 \cr} {/eq}

Hence the stationary point is {eq}\displaystyle x= 0 {/eq}

Now,

{eq}f''(x)=-0.4<0 {/eq}

At the stationary point {eq}\displaystyle x= 0 {/eq}, we have maximum revenue.

Hence the point {eq}\displaystyle x= 0 {/eq} is a point of maxima.