 We are going to take a look at an example of an optimization problem involving the maximization of the volume of a cone. We have here a circle and suppose it measures 10 centimeters in radius. What we want to do is take the circle and create a cone out of it. So if we pinch it in the middle in the center of the circle and just fold it over we can make our cone. And the idea is we want to make a cone as large as possible and we are going to try to find the volume of that largest cone. Wanting then to maximize the volume of this cone, recall that the formula for volume of a cone is one third pi r squared h. If we consider for ourselves a visual of what this might look like, remember that the radius of the circle from which we are going to create the cone is 10. The point here of the cone that really was the center of that circle. So the radius is right there. Thinking ahead we know that in order to maximize this equation we are going to want to take the derivative of it, set it equal to zero so as to find our critical numbers which will therefore locate where it is we hopefully have a relative maximum. The difficulty becomes the fact that our equation is in two variables. The equation as we have it here is what we refer to as our primary equation. Because we are going to need to eliminate one of these variables in order to find our derivative we are going to need a secondary equation. The secondary equation can come from a Pythagorean theorem that is written from the creation of a right triangle inside of this cone. The hypotenuse of that right triangle is 10. Let's refer to the height as h and is the radius of the base of the cone. If we write our Pythagorean theorem then we have h squared plus r squared equals 10 squared. What would be really easy notice how we have an r squared in our primary equation we could simply eliminate that because we know r squared from our Pythagorean theorem is simply 100 minus h squared. So in place of the r squared in our primary equation we are going to substitute 100 minus h squared. In the primary equation then we have the volume equaling 1 third pi times the quantity 100 minus h squared which was the r squared times h. Let's simplify that a little bit and we'll go ahead and distribute both the 1 third pi and the h therefore we get 100 pi over 3 times h minus pi over 3 times h cubed. Now notice our primary equation that we're going to differentiate is in terms of simply one variable. If we go ahead and differentiate we have v prime equal to 100 pi over 3 minus pi h squared. We're going to set that equal to 0 and solve it. Upon cross multiplying we can cancel out our pies. We then have h squared is 100 over 3 which leads to h equaling 10 over root 3. If you prefer a decimal approximation for that that's perfectly fine either one is okay. It would be approximately 5.774. This therefore is our critical number. If there's going to be a maximum this is where it's going to occur. If you are going to opt to use the decimal equivalent of 10 over root 3 please do make sure you're storing this decimal value so that you can use the entire decimal in later calculations. We now need to prove that a maximum is going to occur at that value. We have a couple options. We could do the first derivative test with the number line analysis. We could do the second derivative test. The extreme value theorem would not be an option simply because our domain for this would be in between 0 and 10. Hopefully it makes sense that h could not equal either of these end point values because then we wouldn't have a cone. Therefore because we do not have a closed interval we cannot really use the extreme value theorem. Let's try it first with the first derivative test and I'm going to work this example out for you using both. Simply so you can see how each of them play out in terms of proving that we do have a maximum. With the first derivative test remember we're going to have a number line on which we would have our critical number. It's at this critical number that our derivative equals 0 and into the derivative, let me jot it down up here for us for our reference. If we substitute something less than 10 over root 3 so perhaps we choose 5. Remember the decimal equivalent of this was approximately 5.774. We know we cannot go lower than 0 nor can we go higher than 10. If we substitute for instance 5 into this derivative we do obtain a positive answer. If we substitute something larger than 5.774 so for instance perhaps 6 into this derivative we obtain a negative answer. Since our derivative is changing positive to negative that tells us the original function is changing increasing to decreasing which therefore creates our relative maximum. What if we wanted to prove the critical number we obtained was a relative maximum by using the second derivative test. Remember then we will go ahead and find our second derivative which simply would be working from our first derivative up here negative 2 pi h. We're going to evaluate that second derivative at the critical number we obtained. Now because of the nature of the second derivative negative 2 times pi times this h value that is going to obviously yield a negative answer so think about what that tells us. Since the second derivative at this particular point is less than 0 remember that means we have a concave down segment on the curve and because of that we have a relative maximum. So either way by using the first derivative test or the second you are able to prove that at 10 over square root of 3 we do indeed have a relative maximum. Now we simply have to answer the question which was to find the volume of the largest cone possible. If we return to our volume equation and you'll notice I've picked it up at the point where we had that primary equation in terms of only h and we had distributed the 1 third pi and the h. We want to evaluate this volume at the critical number. Now I'm using 10 over root 3. This is where if you did opt for the decimal approximation as you crunch this out in your calculator you will want to use the variable at which you stored that 5.774 value. Please be sure you're doing that in order to obtain as accurate an answer as possible. If you number crunch that out on your calculator hopefully you arrive at the answer of 403.067 since this is a volume measurement it would be measured in cubic centimeters for the final answer or of course you can do centimeters cubed. Either notation is fine and that would be our final answer as to what the maximum volume would be of this cone if we had started out with a circle that measured 10 centimeters in radius.