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A sales team estimates that the number of new phones they will sell is a function of the price that they set. They estimate that if they set the price at $x$ dollars, they will sell $f(x)=4080-10 x$ phones. Therefore, the company's revenue is $x \cdot(4080-10 x)$.

Find the price $x$ which will maximize the company's revenue.

Select the correct answer below:
174
180
188
198
204
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Step 1 ：The problem is asking for the price that will maximize the company's revenue. The revenue function is given by \(x \cdot(4080-10 x)\).

Step 2 ：To find the maximum of this function, we can take the derivative, set it equal to zero, and solve for \(x\). This will give us the critical points of the function.

Step 3 ：We can then test these points to see which one gives the maximum revenue.

Step 4 ：The derivative of the revenue function is \(4080 - 20x\).

Step 5 ：Setting this equal to zero gives us the critical point \(x = 204\).

Step 6 ：Testing this point, we find that the maximum revenue is $416160.

Step 7 ：Final Answer: The price \(x\) which will maximize the company's revenue is \(\boxed{204}\).