Question

1. A company’s market research team determines the following function for the profit (in thousands of dollars) made on the manufacture and sale of xx thousand graphing calculators: PP(xx) = −20x² + 600x − 3600, 0 ≤ xx ≤ 25.

Answer 1

Explanation:

P(x) = -20x² + 600x - 3600

Where:

P(x) is the profit in thousands of dollars.

x represents the number of thousand graphing calculators manufactured and sold.

The function is defined for 0 ≤ x ≤ 25, which means the number of calculators produced and sold should be between 0 and 25 thousand units.

To maximize profit, you need to find the value of x that maximizes P(x). You can do this by finding the vertex of the quadratic function. The vertex of a quadratic function ax² + bx + c is given by:

x = -b / (2a)

In this case, a = -20 and b = 600:

x = -600 / (2(-20))*

x = -600 / (-40)

x = 15

So, the number of thousand graphing calculators that maximizes profit is x = 15. Now, you can find the maximum profit by plugging this value back into the profit function:

P(15) = -20(15)² + 600(15) - 3600

P(15) = -20(225) + 9000 - 3600

P(15) = -4500 + 9000 - 3600

P(15) = 0

The maximum profit occurs when 15,000 graphing calculators (15 x 1000) are manufactured and sold, and in this case, the maximum profit is $0 (break-even point). If more than 15,000 calculators are produced and sold, the company will start incurring losses.