week six was definately stressful.
.. in the beginning of the week we learned the different theorems.
EVT: a continuous function on a closed interval [a,b] must have both a mnimmum and a maximum on the interval
Rolle's: Let F be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). If f(a)=f(b), then there is at least one number "c" in (a,b) such that f(c)=0.
MVT: If F is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number "c" in (a,b) such that f '(c)=f(b)-f(a)/b-a
and later in the week we recieved packets on optimization..
OPTIMIZATION:
A rancher wants to fence in an area of 3000000 square feet in a rectangular
field and then divide it in half with a fence down the middle parallel to one
side. What is the shortest length of fence that the rancher can use?
There is a top side side and a bottom side (that's two sides).
There are 3 sides the run vertically - one on either side and one in the middle.
Let's call the top and the bottom x.
Let's call the vertical ones y.
The total distance is 2x + 3y.
We know that the area, 3,000,000, is defined by xy.
This says that y = 3000000/x
The equation for length is then 2x + 9000000/x.
There derivative of this function is 2 - 9000000/x², which we set to 0 and solve for x.
That gives us x² = 9000000/2 = 4,500,000.
The answer for x would be √(4.5) thousand.
The answer for y would be 3M/x.
.. i can easly explan both optimization and the theorems, but when it comes to actually applying them i get stuck. i think i have more difficulty with optimization because i get lost in the steps and forget what exactly the questions asking me to find, the only thing i can identify and the primary and secondary functions.. after that i quit. anyone have any ideas on how to make sense of what i'm supposed to do after that?
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I'm probably not the best person to explain this but i will try because everyone's question is on the same thing.
ReplyDeleteyou said you can identify the primary and secondary equations which is step one
for step 2, you solve the secondary equation for one variable if its not already solved for one and then you plug what you get into the primary equation
3. take the derivative of that equation, set equal to zero, and solve for x
4. plug that back into the secondary equation to find the value of your other variable.
thats pretty much the basic steps but i don't think it works for every problem
i think we just all need more practice...but follow these stepss...
ReplyDelete1. Identify all given quantities and quantities to be determined.
2. Write a primary equation for the quantity that you want to max or min
3. Reduce the primary equation so it only has one variable.
4. find the domain of the primary equation.
5. Then find max and mins by derivative set equal to zero.