Answer: C. 8 hundred thousand
Step-by-step explanation:
Assuming the given profit function is:
[tex]P(x) =- x^3+15x^2-48x+450[/tex]
Then we differentiate:
[tex]P'(x)=-3x^2+30x-48[/tex]
And we set the derivative equal to 0 and solve the resulting equation in order to get the critical points:
[tex]-3x^2+30x-48=0[/tex]
Divide by -3 the equation:
[tex]x^2-10x+16=0[/tex]
Factor:
[tex](x-8)(x-2)=0[/tex]
Set each factor equal to zero and solve:
[tex]x-8=0\to x=8[/tex]
[tex]x-2=0\to x=2[/tex]
Then we check which is the maximum by using the first derivative test.
Into the derivative, let’s plug x=0 which is a value in the interval before the critical point x=2:
[tex] P'(0)=-3(0)^2+30(0)-48=-48[/tex]
The derivative is negative, so the profit function is decreasing before x=2
Into the derivative, let’s plug x=3 which is a value in the interval after the critical point x=2 and before the critical point x=8:
[tex] P'(3)=-3(3)^2+30(3)-48=15[/tex]
The derivative is positive, so the profit function is increasing between x=2 and x=8
Notice then at x=2 we have a local minimum since the function changed from decreasing to increasing there.
Into the derivative, let’s now plug x=10 which is a value in the interval after the critical point x=8:
[tex] P'(10)=-3(10)^2+30(10)-48=-48[/tex]
The derivative is negative, so the profit function is decreasing after x=8.
Therefore, at x=8 we have a local maximum because the function changed from increasing to decreasing there.
So, the solution is x=8, meaning 8 hundred thousand tires must be sold to maximize profit. That is option C.
