In calculus, at the maximum point, gradient is zero.
[tex]\begin{gathered} We\text{ plot a graph of the curve P=}-0.02n^2\text{ +}3.40n\text{ - 16}\ldots\ldots.\ldots\ldots\text{.(eqn 1)} \\ \text{Gradient of the curve at any point is }\frac{dP}{dn}=\text{-}0.04n\text{ +3.40} \end{gathered}[/tex]
Recall that at the maximum point, this dP/dn is zero and in that case, we have:
[tex]\begin{gathered} 0\text{ = -0.04n + 3.40 } \\ \text{adding 0.04n to both sides give} \\ 0.04n\text{ = 3.40} \\ \text{dividing both sides by 0.04} \\ \frac{0.04n}{0.04}=\frac{3.40}{0.04} \\ n\text{ = 85 thousand units.} \end{gathered}[/tex][tex]\begin{gathered} \text{Next step is to substitute n into the equation for P (eqn 1)} \\ P=-0.02(85^2)\text{ + 3.4(85) - 16 = 128}.5\text{ thousand dollars.} \end{gathered}[/tex]
Therefore the company will produce 85 thousand units to maximize profits
and in producing these, will make its maximum profit which is 128.5 thousand dollars
