Respuesta :

The revenue function is,

[tex]R(x)=700x-0.5x^2[/tex]

For maximum and minimum value, the first derivative of function is equal to 0.

Determine the first derivative of revenue function.

[tex]\begin{gathered} \frac{d}{dx}R(x)=\frac{d}{dx}(700x-0.5x^2) \\ R^{\prime}(x)=700-0.5\cdot2x \\ =700-x \end{gathered}[/tex]

For maximum and minimum value, R'(x) = 0. So

[tex]\begin{gathered} 0=700-x \\ 700=x \end{gathered}[/tex]

Since x = 700 corresponds to maximum or minimum value of revenue.

Determine the second derivative of revenue function.

[tex]\begin{gathered} \frac{d}{dx}R^{\prime}(x)=\frac{d}{dx}(700-x) \\ R^{\doubleprime}(x)=-1 \end{gathered}[/tex]

Since second derivative of revenue function is less than 0 for every value of x. So x = 700 corresponds to maximum value of revenue.

So 700 villas rented to produce maximum revenue.

Answer: 700 villas

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