Submit test
The equation below is an approximation to the total profit (in thousands of dollars) from the sale of $\mathrm{x}$ hundred thousand tires. Find the number of tires that must be sold to maximize profit.
$P(x)=-x^{3}+\frac{27}{2} x^{2}-60 x+100, x \geq 5$
A. 450,000
B. 550,000
C. 500,000
D. 400,000

Step 1 ：Find the derivative of the function \(P(x)=-x^{3}+\frac{27}{2} x^{2}-60 x+100\), which is \(P'(x)=-3x^{2}+27x-60\).

Step 2 ：Set \(P'(x)\) equal to zero to get the equation \(-3x^{2}+27x-60=0\).

Step 3 ：Simplify this equation by dividing all terms by -3, which gives \(x^{2}-9x+20=0\).

Step 4 ：Factor the equation to get \((x-4)(x-5)=0\).

Step 5 ：Setting each factor equal to zero gives \(x=4\) and \(x=5\).

Step 6 ：Discard \(x=4\) because the problem states that \(x \geq 5\).

Step 7 ：Therefore, the number of tires that must be sold to maximize profit is \(x=5\) hundred thousand, or 500,000 tires.

Step 8 ：Take the second derivative of \(P(x)\), which is \(P''(x)=-6x+27\).

Step 9 ：Substitute \(x=5\) into \(P''(x)\) to get \(-6*5+27=-3\), which is less than zero, confirming that \(x=5\) is a maximum point of the function.

Step 10 ：\(\boxed{500,000}\) is the number of tires that must be sold to maximize profit.