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Difficulty: HardQuadratic Equations

The profit P(x)P(x), in dollars, a company makes from selling xx units of a product is given by the function P(x)=−3x2+bx−1,500P(x) = -3x^2 + bx - 1,500, where bb is a positive constant. If the maximum profit the company can make is 1,2001,200 dollars, what is the value of bb?

Answer: 180

Answer

180
The maximum profit of the quadratic profit function occurs at its vertex. The x-coordinate of the vertex of P(x)=−3x2+bx−1,500P(x) = -3x^2 + bx - 1,500 is given by x=−b2(−3)=b6x = -\frac{b}{2(-3)} = \frac{b}{6}. Substituting this into the profit equation and setting it equal to 1,2001,200 yields −3(b6)2+b(b6)−1,500=1,200-3\left(\frac{b}{6}\right)^2 + b\left(\frac{b}{6}\right) - 1,500 = 1,200. Simplifying this expression results in b212=2,700\frac{b^2}{12} = 2,700, which gives b2=32,400b^2 = 32,400. Since bb is positive, b=180b = 180.

Step-by-Step Solution

1
Identify the x-coordinate of the vertex of the quadratic function in terms of bb.
The x-coordinate of the vertex is x=b6x = \frac{b}{6}.
A quadratic function in the form f(x)=ax2+bx+cf(x) = ax^2 + bx + c has its maximum value at the vertex when the coefficient of the squared term, aa, is negative. The x-coordinate of the vertex is given by x=−b2ax = -\frac{b}{2a}. For this function, a=−3a = -3, so the x-coordinate of the vertex is x=−b2(−3)=b6x = -\frac{b}{2(-3)} = \frac{b}{6}.
2
Substitute x=b6x = \frac{b}{6} back into the profit equation P(x)P(x) and set it equal to the maximum profit of 1,2001,200 dollars.
−3(b6)2+b(b6)−1,500=1,200-3\left(\frac{b}{6}\right)^2 + b\left(\frac{b}{6}\right) - 1,500 = 1,200
The maximum profit of 1,2001,200 dollars is the value of the function at the vertex.
3
Simplify the equation to solve for the positive constant bb.
b=180b = 180
Simplifying the term −3(b6)2-3\left(\frac{b}{6}\right)^2 gives −3(b236)=−b212-3\left(\frac{b^2}{36}\right) = -\frac{b^2}{12}. Simplifying the term b(b6)b\left(\frac{b}{6}\right) gives b26\frac{b^2}{6}. Substituting these back in gives −b212+b26−1,500=1,200-\frac{b^2}{12} + \frac{b^2}{6} - 1,500 = 1,200. Combining the b2b^2 terms gives b212−1,500=1,200\frac{b^2}{12} - 1,500 = 1,200. Adding 1,5001,500 to both sides gives b212=2,700\frac{b^2}{12} = 2,700. Multiplying both sides by 1212 yields b2=32,400b^2 = 32,400. Taking the square root of both sides gives b=±180b = \pm 180. Since bb must be a positive constant, b=180b = 180.

Key Concept

Finding the maximum value of a quadratic function by using the vertex formula x=−b2ax = -\frac{b}{2a} and evaluating the function at that point.
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Topics

Quadratic Equations

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