A company has determined that its weekly profit is a function of the number of items that it sells. Which could represent the weekly profit in thousands of dollars, [tex]$y$[/tex], when the company sells [tex]$x$[/tex] items?
[tex]$y^2=4 x^2-100$[/tex]
[tex]$y=-x^2+50 x-300$[/tex]
[tex]$x=-y^2+60 y-400$[/tex]
[tex]$x^2=-6 y^2+200$[/tex]
Answer (1)
Analyze each equation to determine if it can represent y as a function of x , considering that both x and y are likely non-negative.
Equation 1: y 2 = 4 x 2 − 100 implies x ≥ 5 .
Equation 2: y = − x 2 + 50 x − 300 is a quadratic function with roots approximately x = 6.97 and x = 43.03 .
Equation 3: x = − y 2 + 60 y − 400 implies 0 ≤ x ≤ 500 .
Equation 4: x 2 = − 6 y 2 + 200 implies 0 ≤ x ≤ 200 ≈ 14.14 .
The equation that could represent the weekly profit is y = − x 2 + 50 x − 300 .
Explanation
Understanding the Problem We are given four equations and need to determine which one could represent the weekly profit y (in thousands of dollars) as a function of the number of items sold x . We assume that both x and y are non-negative since the number of items sold and profit cannot be negative.
Analyzing Equation 1 Let's analyze each equation:
Equation 1: y 2 = 4 x 2 − 100 . Solving for y , we get y = ± 4 x 2 − 100 . Since y must be non-negative, we consider only the positive root: y = 4 x 2 − 100 . For y to be real, 4 x 2 − 100 ≥ 0 , which means x 2 ≥ 25 , so x ≥ 5 . This equation could represent the weekly profit as a function of the number of items sold, with the condition that x ≥ 5 .
Analyzing Equation 2 Equation 2: y = − x 2 + 50 x − 300 . This is a quadratic function. To determine if there are non-negative values of x that result in non-negative values of y , we can find the roots of the quadratic equation − x 2 + 50 x − 300 = 0 . Using the quadratic formula or the tool, the roots are approximately x = 6.97 and x = 43.03 . Since the coefficient of x 2 is negative, the parabola opens downward, meaning y is non-negative between the roots. Therefore, for 6.97 ≤ x ≤ 43.03 , y is non-negative. This equation could represent the weekly profit as a function of the number of items sold.
Analyzing Equation 3 Equation 3: x = − y 2 + 60 y − 400 . We need to express y as a function of x . We can rewrite this as y 2 − 60 y + ( 400 + x ) = 0 . Solving for y using the quadratic formula, we get y = 2 60 ± 6 0 2 − 4 ( 400 + x ) = 30 ± 900 − ( 400 + x ) = 30 ± 500 − x . For y to be real, 500 − x ≥ 0 , so x ≤ 500 . Also, y must be non-negative. Since we have y = 30 ± 500 − x , we need to ensure that 30 − 500 − x ≥ 0 . This means 500 − x ≤ 30 , so 500 − x ≤ 900 , which means x ≥ − 400 . Since x must be non-negative, we have 0 ≤ x ≤ 500 . This equation could represent the weekly profit as a function of the number of items sold.
Analyzing Equation 4 Equation 4: x 2 = − 6 y 2 + 200 . Solving for y , we get 6 y 2 = 200 − x 2 , so y 2 = 6 200 − x 2 , and y = ± 6 200 − x 2 . Since y must be non-negative, we consider only the positive root: y = 6 200 − x 2 . For y to be real, 6 200 − x 2 ≥ 0 , which means 200 − x 2 ≥ 0 , so x 2 ≤ 200 , and 0 ≤ x ≤ 200 ≈ 14.14 . This equation could represent the weekly profit as a function of the number of items sold.
Determining the Answer All four equations could represent the weekly profit as a function of the number of items sold, given certain constraints on x and y . However, the question asks which could represent the weekly profit. Since all options are plausible under certain conditions, we need to choose the one that is a function of x. All options are functions of x. Therefore, we choose the simplest one that is a function. The second equation is the most straightforward function of x. Therefore, the answer is y = − x 2 + 50 x − 300 .
Final Answer The equation that could represent the weekly profit in thousands of dollars, y , when the company sells x items is y = − x 2 + 50 x − 300 .
Examples
Understanding the relationship between the number of items sold and the weekly profit is crucial for business planning. For example, a company can use this relationship to determine the optimal number of items to produce and sell in order to maximize profit. By analyzing the equation y = − x 2 + 50 x − 300 , the company can find the number of items that need to be sold to reach the maximum profit. This kind of analysis helps in making informed decisions about production, marketing, and sales strategies, leading to better financial outcomes.
