The demand function for a manufacturer's product is \( p=\displaystyle\frac{60-x}{7} \), where \( x \) is the number of units and \( p \) is the price per unit. At what value of \( x \) will there be maximum revenue?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 30
Revenue is price times quantity:
\( \displaystyle R(x)=x\cdot p=\frac{60x-x^{2}}{7} \)
Setting the derivative to zero:
\( \displaystyle R'(x)=\frac{60-2x}{7}=0 \)
\( x=30 \)
Since \( R''(x)=-\frac{2}{7}\lt 0 \), the revenue is maximum at \( x=30 \).