This week we didn't have much school and i was sick so i only took one of the practice ap
So this is how you do Optimization:
1. Identify primary and secondare equations
(primary is the one you are maximizing or minimizing)
(secondary is the other equation)
2.Solve secondary equation for one variable and plug into the primary
3.Take the derivative of primary equation, set equal to zero, then solve for x.
4. Plug into secondary equation to find the other value.
Example:
Find the length and width of a rectangle with a perimeter of 60 meters and a maximum area.
1. since your maximizing the area, the area of a rectangle formula will be the primary. A=lw
The perimeter formula would be the secondary equation because you give the perimeter 60=2l+2w
2. Solve secondary equation for one variable.
60=2l+2w
60-2w=2l
l=60-2w/2
Now plug into your primary.
A=60-2w/2(w)
3. Multiply the w in to simplify equation.
60w-2w^2/2=30w-w^2
then take the derivateve 30-2w
4. Solve for w.
20=2w
w=15
PLug in you secondary
60=2l+2(15)
60=2l+30
30+2l
l=15
I'm still having trouble with implicit derivatives so if anyone can help me that would be great thanks
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Implicit derivatives are derivatives that involves both x and y.
ReplyDeleteThe steps are:
1. Take the derivative of both sides, same as you would with a normal derivative.
2. Every time you take the derivative of y, note it with dy/dx or y'.s
3. Solve for dy/dx.
Example: y^3 + y^2 -5y -x^2=-4
3y^2 dy/dx + 2y dy/dx - 5 dy/dx - 2x = 0
dy/dx = 2x / 3y^2 + 2y -5
The only extra thing you have to remember is to note the derivative of y with dy/dx and then solve for dy/dx.
Implicit derivatives involve both x's and y's, unlike normal derivatives.
ReplyDelete1: So, first you have to take the derivative of whatever they give you as you normally would.
2: Whenever you take the derivative of y, you have to note it with dy/dx.
3: Solve for dy/dx
(if you want to find slope plug in an x and y value)
example: y^3+y^2-5y-x^2=-4
First you just take the derivative, but don't forget to not the derivatives of the y's! So you get: 3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
Then you have to solve for dy/dx, so you get:
dy/dx(3y^2+2y-5)=2x which then is further solved for to get dy/dx=2x/(3y^2+2y-5)
implicit derivatives have to have both a y and x in order for it to use the implicit derivative rules. so that's the first thing you look for.
ReplyDelete1. take the derivative like you normaly would, and note the derivative of y with dy/dx behind it.
2. simply solve for dy/dx.
pretty easy, just don't forget to note it with dy/dx. because that would screw you up!
Implicit derivatives deal with y and x. It also has an = sign.
ReplyDeleteThe steps are really easy:
1. take derivative of both sides
*whenever taking the derivative of y note behind it with dy/dx
2. solve for dy/dx
*move everything with dy/dx to one side and the other stuff to the other side, divide by dy/dx