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Difficulty: MediumQuadratic Equations

The daily profit, in dollars, of a small company is modeled by the quadratic function P(x)=−2x2+120x−1000P(x) = -2x^2 + 120x - 1000, where xx represents the number of items the company produces and sells each day. For what number of items produced and sold, greater than 2020, will the company break even (meaning its daily profit is $0\$0)?

  1. A
    10
  2. B
    30
  3. C
    40
  4. 50Answer

Answer

The company breaks even when it produces and sells 50 items.
To find the break-even points, set the profit function to zero: −2x2+120x−1000=0-2x^2 + 120x - 1000 = 0. Dividing both sides by −2-2 yields x2−60x+500=0x^2 - 60x + 500 = 0. Factoring this equation gives (x−10)(x−50)=0(x - 10)(x - 50) = 0, which results in the solutions x=10x = 10 and x=50x = 50. Since the question specifies that the number of items must be greater than 2020, the correct answer is 50.

Step-by-Step Solution

1
Set the profit function equal to zero to find the break-even points.
−2x2+120x−1000=0-2x^2 + 120x - 1000 = 0
Breaking even means the profit, P(x)P(x), is equal to 0.
2
Divide the entire equation by the common factor of −2-2 to simplify the quadratic expression.
x2−60x+500=0x^2 - 60x + 500 = 0
Simplifying the quadratic expression makes it easier to factor.
3
Factor the quadratic equation by finding two numbers that multiply to 500500 and add to −60-60.
(x−10)(x−50)=0(x - 10)(x - 50) = 0
Factoring allows us to find the individual roots of the equation.
4
Solve for xx by setting each factor equal to zero.
x=10x = 10 or x=50x = 50
The zero product property states that if a product of factors is zero, at least one factor must be zero.
5
Apply the constraint that the number of items must be greater than 2020.
x=50x = 50
Out of the two roots, only 50 is greater than 20.

Key Concept

Solving quadratic equations in real-world contexts by finding roots and applying constraints.
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Topics

Quadratic Equations

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