Linda owns a jeans store. Her montly profit p( x) dollars on the sale of x pairs of jeans is given by the equation p(x ) = 50x - 1/5x^2. How many pairs of jeans must she sell each month to maximize her profit?

Take the derivative of the equation, set it equal to zero, and solve for x.

To maximize Linda's profit, we need to find the value of x that corresponds to the maximum point on the profit curve.

First, we need to determine the derivative of the profit function p(x) with respect to x. The derivative will give us the rate of change of the profit function.

p(x) = 50x - 1/5x^2

To find the derivative, we differentiate each term separately:

p'(x) = 50 - (2/5)x

Setting p'(x) equal to zero, we can find the critical points:

50 - (2/5)x = 0
(2/5)x = 50
x = (5/2) * 50
x = 125

So, the critical point is x = 125.

Now, we are looking for the maximum point on the profit curve. We check the second derivative to determine if the critical point is a maximum or minimum.

To find the second derivative, we differentiate p'(x):

p''(x) = -2/5

Since the second derivative is negative, the critical point x = 125 corresponds to a maximum point on the profit curve.

Therefore, Linda must sell 125 pairs of jeans each month to maximize her profit.