Week 7

Calculus Week#7

So this week in Calculus we started off by studying and doing more problems with Optimization. The thing about Calculus in general that I've learned is that everything comes with more and more practice. We took a short quiz on optimization which I thought was pretty easy, but I know a lot of you struggled with it...and I hope that won't happen on the exam or something.

To go over optimization, the way I think about it:

You look at the problem. Read what it says...if it says to find the "maximum area" or the "area of maximum size", it obviously means you are going to be optimizing the area...because it wants the maximum..or minimum.

So, we know our primary equation. Now we look at the equation. Does it have 3 or more variables? If so, you have to find a secondary equation. In the problem it will probably say something like "Find the maximum area of a rectangle that has a perimeter of 36 units". Okay, so perimeter was mentioned...let's make that our secondary.

So now you have two equations and want to maximize one. So, to make the one that you want to maximize in terms of one variable, solve the secondary equation for a variable and then plug in what you get for that variable in the primary.

For instance, say we have

2x+2y=36 and xy=A

Solving the secondary (the first one) for y gives us (36-2x)/2. Simplifying and plugging into the first one for y, you get x(18-x)=A.

Now that we have it in terms of one variable, simplify and take the derivative.

(18x-x²).
The derivative is simply 18-2x.

Now we set the derivative equal to 0 and solve. x=9.

Plugging that back into our secondary, which is the next step, would give us that y=18 as well.

You are done. :-) The maximum area rectangle has dimensions 9x9. Btw, 9+9+9+9 = 36 (the perimeter works out if you want to check yourself)

Also, a hint for the future of rectangles....the maximum area will always be a n x n sized rectangle...that's just how it works.

Good luck on study guides and exam preparation!