"A rectangular garden is to be constructed with one side against the side of a barn. If 60 meters of fencing are available to enclose the other three sides of the garden, what dimensions should be used to maximize the area of the garden?"

To solve this problem, we need to use the concept of optimization in calculus. Here are the steps:

1. Define the variables: Let's denote the length of the garden parallel to the barn as x and the width of the garden (the sides perpendicular to the barn) as y.

2. Write down the constraints: We know that the total length of the fence is 60 meters, and it's used to fence two sides of width y and one side of length x. So, we have the equation 2y + x = 60.

3. Express the quantity to be maximized or minimized as a function of one variable: We want to maximize the area A of the rectangle, which is A = xy. From the constraint equation, we can express y as y = (60 - x) / 2. Substituting this into the area function gives A = x(60 - x) / 2 = 30x - x^2/2.

4. Find the critical points of the function: To find the critical points, we take the derivative of A with respect to x and set it equal to zero. The derivative dA/dx = 30 - x. Setting this equal to zero gives x = 30.

5. Determine the maximum or minimum value: We can see that the second derivative of A with respect to x is -1, which is less than zero. So, by the second derivative test, x = 30 is a maximum point.

6. Answer the problem: Substituting x = 30 into the constraint equation gives y = (60 - 30) / 2 = 15. So, the dimensions that maximize the area of the garden are 30 meters parallel to the barn and 15 meters perpendicular to the barn.