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. Based on this model:

a. Find the y-intercept and explain what it means in this context.

b. Find the x-intercepts and explain what they mean in this context.

c. How many cars should be made and sold to maximize profit?

d. What is the maximum profit?

The y intercept is - $66,000

The x intercept is 20 and 330 cars

175 cars would be sold to make the maximum profit

Maximum profit is $240250

Step-by-step explanation:

From the equation

P (x) = -10x²+3500x-66000

The y values are values for the profits made from sales of vehicles or p (x)

While the x values are the values for number of vehicles sold.

To obtain the y intercept i. e. the profit that will be accrued when no vehicle is sold. This is obtained by putting x=0 in the equation above. This will yield y=-$66000

Meaning the company will b running at loss if not in operation

The x intercepts indicated the number of vehicles that needed to be sold so thay there won't be either profit or loss. Mathimatically, it means what will be the values of x if y is 0.

10x²-3500x+66000=0

Solving this quadratic equation, we find x to be 20 or 330 vehicles. When 20 or 330 vehicles are produced, there won't be profit or loss.

From the equation, we can see that our graph (parabola) will open downwards due to the - 10 in the coefficient of x².

To obtain the number of vehicles at the vertex (maximum profit), we use - b/2a

By comparing our equation with

y=ax²+bx+c

b=3500

a=-10

-b/2a=-3500/-10=175

The number of vehicles that will yield the maximum profit is 175 cars

Since the number of cars (175) to make the maximum profit is know, the maximum profit will be obtained from

P (x) = -10 (175) ²+3500 (175) - 66000

P (x) = $240250