Reflection

This week in calculus, we learned how to solve optimization problems.
STEPS:
1. Identify all given quantities and what you have to find.
2. Find a primary equation.
3. Find a secondary equation if necessary. (Only if there are 2 variables in
equation in step 2.)
4. Find max or min by plugging into primary equation doing first derivative test.
Instead of intervals plug back into primary equation to find highest and lowest
value.

EXAMPLE from notes:
Find two numbers whose sum is 10 and product is as large as possible.
1. a+b=10
2. P=ab > primary
3. a+b=10
b=10-a
4. P=a(10-a)
=10a-a^2
Derivative: 10-2a=0
a=5
5 > 10(5)-5^2=25
P=ab
25=5(b)
b=5

EXAMPLE from quiz:
A farmer has 2400 ft of fencing. What are the dimensions of a large area of the pen.
1. P=2400
l>b
w>a
2. A=ab
3. 2a+2b=2400
4. 2a=2400-2b
a=1200-b
A=(1200-b)b
=1200-b^2
1200-2b=0
-2b=-1200
b=600
A=1200(600)-600^2
A=3600
A=ab
3600=600a
a=6000
> 600 X 600