The profit from selling x pieces of handmade jewelry can be modeled by the equation P(x)=-x^2+130x-3000. What is the range of profitable orders
(P(x)>0)?
find the vertex by whatever means you have learned.
so the range is between 0 and the y of your vertex
that doesnt help
To find the range of profitable orders, we need to determine the values of x for which P(x) is greater than zero (P(x) > 0). In other words, we want to find the values of x that yield a positive profit.
The given equation is P(x) = -x^2 + 130x - 3000.
To find the range of profitable orders, we need to solve the inequality -x^2 + 130x - 3000 > 0.
Step 1: Set up the inequality:
-x^2 + 130x - 3000 > 0
Step 2: Solve the inequality:
To solve the inequality, we can factor the quadratic equation or use the quadratic formula. In this case, let's use the quadratic formula to find the roots of the equation:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation -x^2 + 130x - 3000 > 0, the coefficients are:
a = -1
b = 130
c = -3000
Plugging in these values into the quadratic formula:
x = (-130 ± √(130^2 - 4(-1)(-3000))) / 2(-1)
Simplifying further:
x = (-130 ± √(16900 - 12000)) / -2
x = (-130 ± √4900) / -2
x = (-130 ± 70) / -2
Now we have two possible solutions for x:
x1 = (-130 + 70) / -2 = -60 / -2 = 30
x2 = (-130 - 70) / -2 = -200 / -2 = 100
Step 3: Determine the profitable range:
Now that we have the two possible values of x, we need to determine which range satisfies the inequality -x^2 + 130x - 3000 > 0.
Since the coefficient of x^2 is negative (-1), the parabola opens downward. The graph of this parabola will look like an upside-down U. To satisfy the inequality, the graph needs to be positive (above the x-axis).
Plugging in a test value from each interval into the inequality, we can determine the ranges that yield positive results:
For x < 30, we can choose a value less than 30, like x = 0:
P(0) = -(0^2) + 130(0) - 3000 = 0 - 0 - 3000 = -3000 < 0
For 30 < x < 100, we can choose a value between 30 and 100, like x = 50:
P(50) = -(50^2) + 130(50) - 3000 = -2500 + 6500 - 3000 = 1000 > 0
For x > 100, we can choose a value greater than 100, like x = 150:
P(150) = -(150^2) + 130(150) - 3000 = -22500 + 19500 - 3000 = -3000 < 0
Therefore, the profitable range of orders (P(x) > 0) is 30 < x < 100.