Marginal revenue and maximum revenue

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If the price at which $x$ units of a product sell is $p(x)=50-0.1x$ (euros), the revenue is $R(x)=x\,p(x)$. Compute the marginal revenue $R'(x)$ at $x=100$ and find the quantity that maximizes revenue, with the corresponding revenue.

Solution

$R(x)=50x-0.1x^{2}$, so $R'(x)=50-0.2x$. Marginal revenue at $x=100$: $R'(100)=50-20=30$ euros/unit. Maximum: $R'(x)=0\Rightarrow x=250$; $R(250)=12500-6250=6250$ euros. $\boxed{R'(100)=30;\quad x_{\max}=250;\quad R_{\max}=6250\ \text{euros}}$