We have learned several derivative techniques. We know that derivatives give us the slope of a curve at any instant (where the function is differentiable). If we then look for times when the slope is zero, we will know when the curve reaches at least a local maximum or local minimum. To be able to find maxima and minima is an extremely powerful ability. This lesson explores this ability. We will be able to solve optimization problems. We can actually find optimal ways of doing things. Just imagine the power of a businessman who can find how to minimuze costs and maximize profit! Imagine the importance of maximizing health quality! Imagine the wonderful byproducts of minmizing our damage to the environment! Imagine the safety, speed, efficiency of fuel consuption when optimizing is applied to studying flight efficiency! After this lesson you will be shown the presentations "Optimal Design," and "Thinking God's Thoughts after Him," both from the series: "Divine Design" which will explore many of these issues.
Probably the biggest challenge in solving optimization problems is setting the problem up. This almost always requires a diagram, setting up variables, and finding and writing the mathematical relationship between the variables. Once this setup is complete, then calculating the derivative and setting that derivative equal to zero can lead to optimal solutions.
Sample 1
Find the dimensions of a rectangle with maximum area if the perimiter must be 1000 meters.  Given Problem. 
First, we make the diagram of a rectangle. We also begin making variable choices by setting one side of the rectangle equal to x.  
Now, instead of naming the length of the rectangle with a seperate variable, we note that we had 1000  
Now we express the area (which is what was to be maximized) as a function of x.  
Because we want to maximize the area, we take the derivative.  
We then set the derivative equal to zero, so that we can find any critical points. We solve this equation. Because A(x) is downward opening parabola, we are assured that our critical point of x = 250 is a maximum. So we now know that a square 250 x 250 will give us the maximum area if the perimeter is fixed at 1000. 
Example 2
A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have.  Given Problem, (#8 in text). (This is a classic problem type found in most calculus textbooks) 
We make a drawing of the situation.We have our 3 ft square piece of cardboard, and we have called the dimensions of the cutout squares x by x. 

If we "cut out" the square corners and sketch in some lines representing the base of the new box we have this situation. Notice the new dimension for the length and width of the base (32x).  
To find the volume of a rectangular prism or box, we need to multiply the area of the base times the height.  
After multiplying, we arrive at our volume function in terms of x.  
Now we take the derivative of the function.  
We set the derivative equal to zero and solve. Both answers we got make algebra sense. However 1.5 can't be an answer, otherwise the (32x) dimension would be zero, and there would be no bottom to our box, and therefore no volume.  
Now we take our optimum value for x and calculate the volume. 
Using your Tools for Enriching Calculus CD (that came with your book), load and run Module 4.6. This module takes you through eight additional optimization problems, including animations of the physical situations. 
If you have the Journey Through Calculus CD, load and run Resources/Module 5/Max and Min/Start of Optimal Lifeguard 
If you have the Journey Through Calculus CD, load and run Resources/Module 5/Max and Min/Start of Max and Min 
