# Applications of Derivatives: Optimization

Applications of Derivative: Optimization Problems

Here we’ll look at how to use the derivative to solve optimization problems or ‘minimum/maximum’ problems.

Say we have an equation of power output versus time for a given system. It might be useful to know when the maximum power output occurs.

p(t) = -0.5t3 + 2.5t – 0.2

Let’s look at the plot of this curve.

NEED SKETCH

We can easily see where the maximum power occurs on the plot. What is the slope of the curve at the maximum (or minimum) point?

Zero, right?

And how do we determine the slope of a curve?

Take the derivative. Great we already know how to do that. Let’s try it.

dp/dt = -1.5t2 + 2.5

Now set this equal to 0.

0 = -1.5t2 + 2.5

t = +/- 1.29 seconds

Negative time doesn’t make sense for this problem, so the maximum power output occurs after 1.29 seconds. Plug this back into the original equation:

p(1.29) = -0.5(1.29)3 + 2.5(1.29) – 0.2 = 1.95W

Sometimes we’re not sure if we/ve located a minimum or a maximum point. Then we can look at the second derivative. If it’s positive the curve is concave downward at that point: a minimum. If it’s negative the curve is concave upward at that point: a maximum. For this problem:

d2p/dt2 = -3.0t = -3.87

This is negative, so it IS a maximum point.