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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.
$P(x)=-x^3+\frac{27}{2}{x}^2-60x+100,x\ge 5$
A. 450,000
B. 550,000
C. 500,000
D. 400,000

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

Step 2 ：Set $P^{\prime }(x)$ 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 $x\ge 5$.

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 $-6\ast 5+27=-3$, which is less than zero, confirming that $x=5$ is a maximum point of the function.

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