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An automobile manufacturer can produce up to 300 cars per day. The profit made from the sale of these vehicles can be modeled by the function:

\[ P(x) = -10x^2 + 3500x - 66,000 \]

where \( P(x) \) is the profit in dollars and \( x \) is the number of automobiles made and sold. How many cars should be made and sold to maximize profit?

Answer :

Final answer:

To maximize profit, find the x-coordinate of the vertex of the quadratic function.

Explanation:

To maximize profit, we need to find the value of x that corresponds to the vertex of the quadratic function.

The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a, b, and c are the coefficients of the quadratic function in the form of ax^2 + bx + c.

In this case, a = -10, b = 3500, and c = -66000.

Plugging in these values, we get

x = -3500 / (2*(-10))

= -3500 / (-20)

= 175.

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Rewritten by : Barada

Answer: 175

Step-by-step explanation:

To find maximum, we need to find the vertex.

The x-value of the vertex is the number of automobiles that should be made and sold. The y-value is the maximum profit.

P(x) = -10x² + 3500x - 66,000

a=-10, b = 3500, c=-66,000

[tex]x=\dfrac{-b}{2a}[/tex]

[tex]x=\dfrac{-(3500)}{2(-10)}[/tex]

[tex]x = \dfrac{-3500}{-20}[/tex]

[tex]x = 175[/tex]


BONUS:

The maximum profit is: P(x) = -10(175)² + 3500(175) - 66,000

= -300,625 + 612,500 - 66,000

= 245,875