Question 23275
Maximise: {{{P = -25x^2 + 300x}}}

This is the equation of a parabola that opens downwards (coefficient of x^2 is negative) so the maximum value of P (the dependent variable) will be found at the parabola's vertex. The x-coordinate of the vertex is given by:
{{{x = -b/2a}}} and so the maximum value of P will be found at {{{x = -b/2a}}}

Your equation is already in the standard form: {{{P = ax^2 + bx + c}}} (a = -25, b = 300, c = 0) so we can find the x-coordinate at which P will be the maximum.

{{{x = -(300)/2(-25)}}} Simplify.
{{{x = -300/(-50)}}}
{{{x = 6}}}

To maximise profits, the manager should employ 6 clerks.

The maximum profit can be found by substituting 6 for x in the original equation for P.

{{{P = -25(6)^2 + 300(6)}}}
{{{P = -25(36) + 1800}}}
{{{P = -900 + 1800}}}
{{{P = 900}}}

Maximum profit is 900 (dollars ?)