𝑝(𝑥) = −2𝑥^2 + 200𝑥 + 25,000

Question

A company that produces cellular phones analyzes its production and finds that the profit P (in dollars) on the sale of the phones is approximated by the function. where x is the number of phones produced each month. How many phones per month should be made so as to maximize profit? And what is the maximum profit?

Answer 1

Create a table relating values of x with P(x) = -2x² +200x + 25000

x P(x)

0 25000

1 -2(1)² + 200(1) +25000 = -2 + 200 +25000 = 25198

2 -2(2)² + 200(2) +25000 = -8 + 400 +25000 = 25392

3 -2(3)² + 200(3) +25000 = -18 + 600 + 25000 = 25582

Notice that as x increases, so does the P(x)

Also notice that the rate of increase in P(x) decreases as x increases.

This means that eventually P(x) will reach a value that will no longer increase for an increase in x. That will be its maximum value.

Let x = 49. Then P(x) = -2(49)² + 200(49) + 25000 = -4802 + 9800 + 25000 = 29898

Let x = 50. Then P(x) = -2(50)² + 200(50) + 25000 = -5000 + 10000 + 25000 = 30000.

Let x = 51. Then P(x) = -2(51)² + 200(51) + 25000 = -5202 + 10200 + 25000 = 29998.

Thus x = 50 is the point where the parabola reaches it's maximum. Any value of x less than 50 (including negative values) or any value of x greater than 50 will result in P(x) being less than its value when x=50.

A more precise way to find the point of inflection is to take the derivative as was done in the first answer.

Answer 2

Given a profit function of P(x) = -2x² + 200x + 25000

This is a quadratic equation so the graph will be a parabola.

To find the maximum (or minimum) we can differentiate the function.

If y = -2x² + 200x + 25000

then ^dy/_dx = -4x + 200

As there's only one max/min and the question asks about maximum then I'd assume this is the maximum. Note: we know by differentiating a second time (-4) and a -ve number denote a maximum.

dy/dx = gradient of tangent to function AND at max/min gradient = 0 (horizontal)

hence 0 = -4x + 200 => x = 50

then y = -2(50)² + 200(50) + 25000 = 30000

Hence the vertex of the parabola is (50, 30000) and the maximum profit is $30,000 which occurs when 50 phones are made.

Differentiation is a general way of finding maxima and minima for different types of functions, not just quadratics. If you haven't done differentiation yet, you can also use the vertex form of a parabola:

a(x - h)² + k where the vertex is at (h, k)

=> a(x² - 2hx + h²) + k => ax² - 2ahx + ah² + k

compare this with -2x² + 200x + 25000

and we see the x² term ax² = -2x² => a = -2

and the x term -2ahx = 200x => 4hx = 200x => h = 50

and the number term ah² + k = 25000 => (-2)(50)² + k = 25000 => -5000 + k = 25000 => k = 30000

hence P(x) can also be written as -2(x - 50)² + 30000 with vertex at (50, 30000)

You can double check with Desmos Graphing or similar and/or online differentiation calculators.