Applications of Derivatives: Optimization

Applications of Derivatives: Optimization

Example 1: You have 100meters of fence to make two adjacent animal pens. What are the dimensions which give the maximum area?

What is it you want to optimize?

    Maximize area, A.

What are you solving for?

    Dimensions, say x and y. We need a sketch to label them.


What is the equation for area?

    A = xy

We need an equation with only one variable, say x. Can we find another equation?

    The perimeter, P = 100 = 2x + 3y

    Solve for y:     y = (100-2x)/3

    Put this into the area equation:

    A = x(100-2x)/3

Now take the derivative and set it equal to zero. First multiply it out:

    A = (100x – 2x2)/3

    dA/dx = (100-4x)/3

    0 = (100-4x)/3

    4x = 100

    x = 25m

    y = 16.7m

Check to make sure this is a maximum:

    d2A/dx2 = -4/3 which tells us concave down–a maximum.