Finding Local Extrema with Derivatives

[scorpa]

Hey there!

You guys aren't going to believe that I can get stuck on something this pathetically easy, but I have to ask anyway.

Find two positive numbers whose product is 108 and the sum of the first number plus three times the second number is a minimum.

Here is what I've done:

xy = 108
x + 3y = S

Let x and y be the two numbers, and S the sum.

To find S substitute x = (108/y) into x + 3y = S

S = (108/y) + 3y

Now I need to take the derivative of the function, and set it equal to zero to find a critical number, but I can't get the right answer for this part, which is stupid because it should be so darn easy. I haven't done derivatives in awhile and now I am starting to forget them

This is what the book says :

dS/dy = -(108/y^2) + 3 = 0

3 = 108/y^2
y^2 = 36
y = 6

The second derivative the book says is 216/y^3

Can someone please explain how they got the derivatives to me? I feel absolutely stupid having to ask this but I figured I had better ask now so I understand it later. Thanks so much

 

[dextercioby]

What are the derivatives of a constant,of $y=x$ and $y=\frac{1}{x}$...

U need them all.

Daniel.

 

[BobG]

S = (108/y) + 3y

I'd rewrite this as $$S = 108y^{-1} + 3y$$

You use the power rule to find the derivative of each term. This identifies your critical points, but doesn't tell you for sure whether you have a local minimum or a local maximum.

Use the power rule, again, to find the derivative of your derivative (the second derivative). If the second derivative is greater than 0, you have a local minimum; If less than 0, a local maximum.

If you don't understand why, then graph your original function, your first derivative, and your second derivative. The relationship between them should be a little clearer.