Q.A firm's total revenue function is $R(x) = 40x - x^2$, where $x$ is the number of units sold.

Write R(x) in the standard quadratic form to identify a, b, c; the vertex formula gives the maximum, and factorising gives the zeros. …

1. Standard form. $R(x) = -x^2 + 40x + 0$ is of the form $ax^2+bx+c$ with $a=-1$, $b=40$, $c=0$ — since $a \neq 0$, $R(x)$ is a quadratic function of $x$. As $a<0$, its graph is a downward-opening parabola, so it has a maximum (not a minimum).
2. Output at which revenue is maximum.

   $x = -\frac{b}{2a} = -\frac{40}{2(-1)} = -\frac{40}{-2} = 20 \text{ units}$

   Maximum revenue:

   $R(20) = 40(20) - (20)^2 = 800-400 = ₹400$

   (iii) Values of $x$ for which revenue is zero. $40x - x^2 = 0 \Rightarrow x(40-x) = 0 \Rightarrow x=0 \ \text{or} \ x=40$ …

Alternative check for part (ii): since the roots are x = 0 and x = 40 (found in part iii), the vertex must lie exactly midway between them, at x = (0+40)/2 = …

Using the formula x = −b/2a but making a sign error since a itself is negative here (a common slip: −40/2(1) instead of −40/2(−1)). Also, stopping after finding x = …