P(x)= -x^3+15x^2-48x+450, x>/3 is an approximation to the total profit (in thousands of dollars) from the sale of x hundred thousand tires. find the number of hundred thousands of tires that must be sold to maximize profit.
0=-3x^2+30x-48
0=x^2-10x+16
0=(x-8)(x-2)
x=8, or x=2
check that.
To find the number of hundred thousands of tires that must be sold to maximize profit, we need to find the critical points of the profit function.
The profit function P(x) is given by P(x) = -x^3 + 15x^2 - 48x + 450.
To find the critical points, we need to take the derivative of the profit function:
P'(x) = -3x^2 + 30x - 48.
Setting P'(x) = 0 and solving for x, we can find the critical points:
-3x^2 + 30x - 48 = 0.
Dividing both sides by -3 gives us:
x^2 - 10x + 16 = 0.
Factoring the quadratic equation, we get:
(x - 2)(x - 8) = 0.
So, x = 2 and x = 8 are the critical points.
Now, we need to determine which of these critical points yields the maximum profit. To do this, we can check the concavity of the profit function by finding the second derivative:
P''(x) = -6x + 30.
Substituting x = 2 and x = 8 into P''(x), we can determine the concavity:
P''(2) = -6(2) + 30 = 18,
P''(8) = -6(8) + 30 = -18.
Since P''(2) > 0 and P''(8) < 0, we conclude that x = 2 corresponds to a local minimum, while x = 8 corresponds to a local maximum.
Therefore, the number of hundred thousands of tires that must be sold to maximize profit is 8.
To find the number of hundred thousands of tires that must be sold to maximize profit, we need to find the maximum point of the profit function.
Given the profit function P(x) = -x^3 + 15x^2 - 48x + 450, we need to find the value of x that maximizes this function.
To find the maximum point, we can use calculus. We need to find the derivative of the profit function, set it equal to zero, and solve for x.
Step 1: Take the derivative of P(x) with respect to x:
P'(x) = -3x^2 + 30x - 48
Step 2: Set P'(x) = 0 and solve for x:
-3x^2 + 30x - 48 = 0
Step 3: Solve the quadratic equation. We can factor out -3 from the equation:
-3(x^2 - 10x + 16) = 0
Now we can solve the quadratic inside the parentheses:
(x - 2)(x - 8) = 0
Setting each factor equal to zero, we find two possible values for x:
x - 2 = 0 or x - 8 = 0
Solving these equations, we get:
x = 2 or x = 8
Step 4: Determine which value of x maximizes the profit function.
To determine if x = 2 or x = 8 maximizes the profit function, we can use the second derivative test. It involves taking the second derivative of the function and evaluating it at the critical points.
Step 4a: Take the second derivative of P(x):
P''(x) = -6x + 30
Step 4b: Evaluate P''(x) at x = 2 and x = 8:
P''(2) = -6(2) + 30 = 18
P''(8) = -6(8) + 30 = -18
Since P''(2) > 0 and P''(8) < 0, the second derivative test tells us that the point x = 2 is a local minimum and x = 8 is a local maximum.
Therefore, to maximize profit, the number of hundred thousand tires that must be sold is x = 8.