Q.Find the maximum profit that a company can make, if the profit function is given by $p(x)=41-72x-18x^2$

The key idea is that a quadratic profit function with a negative coefficient on $x^2$ has a maximum at its vertex.

Step 1: The profit function is $p(x) = 41 - 72x - 18x^2$. This is a downward-opening parabola (coefficient of $x^2$ is $-18 < 0$), so the vertex gives the maximum profit.

Step 2: For a quadratic $ax^2 + bx + c$, the $x$-coordinate of the vertex is $x = -\frac{b}{2a}$. Here $a = -18$, $b = -72$, so: …

The key insight here is that the profit function is a quadratic — a parabola. When the coefficient of $x^2$ is negative, the parabola opens downward, meaning it has a single highest point (a maximum) and no minimum. In business problems, this shape is common: profit often rises to a peak at some optimal production level, then falls.

For any quadratic $ax^2+bx+c$, the vertex (where the maximum or minimum occurs) is at $x = -\frac{b}{2a}$. This formula comes from calculus (setting the derivative to zero) or from completing the square — either way, it’s the exact point where the parabola turns around.

Let’s apply it step by step.

1. Identify the coefficients. The profit function is $p(x)=41-72x-18x^2$. Rewrite it in standard form:

So $a = -18$, $b = -72$, $c = 41$.

2. Find the vertex’s x-coordinate. Using $x = -\frac{b}{2a}$:

The maximum profit occurs at $x = -2$.

Method: Maximizing a Quadratic Cost/Profit Function via the Vertex

Business-context "find the maximum profit/revenue/cost" questions are almost always a quadratic in disguise — the method is identical to finding a parabola's vertex, just applied to a modelling context.

Steps

Step 1: Confirm the profit (or revenue/cost) function is genuinely quadratic and identify $a$, $b$, $c$.

Write the function in standard form $p(x) = ax^2+bx+c$, being careful to correctly identify the sign of each coefficient — profit functions often list terms in a scrambled order (constant first, then a subtraction, then the squared term).

Step 2: Check the sign of $a$ to confirm a maximum genuinely exists.

If instead $a>0$, the vertex would give a minimum, and the "maximum profit" would be unbounded — always confirm this sign before proceeding.

Step 3: Compute the vertex's $x$-coordinate. …

Common Mistakes

Mistake 1: Losing a sign when substituting negative $a$ and $b$ into the vertex formula.

Why it's wrong: with $a=-18$ and $b=-72$, a careless student computes $x=-\frac{-72}{2(-18)}$ and mishandles the double negative, landing on $x=2$ instead of the correct $x=-2$. Correct approach: substitute the signed values in one deliberate step, simplifying $-\frac{(-72)}{-36}$ carefully rather than cancelling signs by eye.

Mistake 2: Not checking that the coefficient of $x^2$ is negative before calling the vertex a "maximum".

Why it's wrong: if a student skips confirming $a<0$ and blindly calls the vertex value the maximum profit, the conclusion would be wrong for any profit function that instead opens upward (which would have no maximum at all). Correct approach: state explicitly that $a=-18<0$, so the parabola opens downward and the vertex is genuinely the maximum. …