Thursday, November 29, 2007

Optimization Problems

Well i am finally back to scribe, it has been a pretty long time since i had last scribed so here it is. Today in class, Mr. K introduced a new topic Optimization problems. Optimization problems are problems that deal with either finding the maximum or minimum of the function in the problem. although very rarely is the function given so you as a calculus student must use all your knowledge you've learnt up to now and create a formula/function that can be used to answer the question that is being asked. With that said Mr. K gave the class the 6 rules to live by when doing optimization problems, although do not really live by them, just follow them loosely.

6 Rules to Optimization Problems

1. Find what you are trying to either maximize or minimize.

2. Write an equation for it, make sure to use descriptive variables so you won't get confused when doing the problem.

3. Try to get the equation into a two variable form, where you are finding one of the variables, because you want to find the derivative.

4. Write the derivative of the equation in the two variable form.

5. Set the derivative to zero, and solve for the remaining values. (First Derivative Test or Second derivative Test)

6. If needed depending on what the question asks, Plug the value of that variable into the one or (two) equations to find all dimensions.

Now here is the explanation of the question done in class.

Well first we identified it was a maximum problems, because it asks for the box with dimensions with the largest volume.

Next the equation is made by using V=h*w*L, so h=x, w=16-2x, and L=21-2x. So the equation looks something like this, x(16-2x)(21-2x).

Then we multiply it out. 336x - 74x^2 + 4x^3

Then we take the derivative. Which is 336 - 148x + 12x^2

then by using the quadratic equation we find that x = 3 and 9.3, so we do a line graph and find it is positive before 3 and negative after 3 meaning that by the use of first derivative test, 3 is the maximum x can be.

Well, that was it for today's class. Tomorrow's Scribe will be M@rk.


Anonymous said...

How do you find the domain?

dkuropatwa said...

Look at the picture on slide #4 above. The domain has to be the values of x that "make sense."

In the case, the smallest possible value of x is zero (which you wouldn't actually do because the volume would be zero, but, it can happen). The largest possible value of x is half the length of the smallest side; in this case 16/2 or 8.

Look at the picture on slide #4 until this makes sense.