A firm's profit function is $P(x) = -2x^2 + 120x - 800$, where $x$ is the number of units sold. Find the number of units that maximises profit, and the maximum profit.

Question

Accepted Answer

$P(x) = -2x^2+120x-800$ is a quadratic function with leading coefficient $a=-2<0$, so its graph is a downward-opening parabola — meaning it does have a single highest point (a maximum), located at its vertex.

**Method — completing the square.** Factor $-2$ out of the first two terms:
$$P(x) = -2\big(x^2 - 60x\big) - 800$$
To complete the square inside the bracket, take half of $60$ (which is $30$) and square it (which is $900$):
$$P(x) = -2\big[(x-30)^2 - 900\big] - 800 = -2(x-30)^2 + 1800 - 800$$
$$P(x) = -2(x-30)^2 + 1000$$

Since $-2(x-30)^2 \le 0$ for every real $x$ (a square is never negative, and it is multiplied by $-2$), the largest possible value of $P(x)$ occurs when $(x-30)^2 = 0$, i.e. at $x=30$, giving the maximum value $P(30) = 1000$. …

Q.A firm's profit function is $P(x) = -2x^2 + 120x - 800$ , where $x$ is the number of units sold. Find the number of units that maximises profit, and the maximum profit.

Concept understanding — Business Functions

Q.A firm's profit function is P(x)=−2x2+120x−800P(x) = -2x^2 + 120x - 800P(x)=−2x2+120x−800, where xxx is the number of units sold. Find the number of units that maximises profit, and the maximum profit.

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Q.A firm's profit function is $P(x) = -2x^2 + 120x - 800$ , where $x$ is the number of units sold. Find the number of units that maximises profit, and the maximum profit.

Unlock everything free for 14 days