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river optimization problem
1500 meters of fencing divided pasture into two parts that runs along the river and you wish to find the max possible area. What are the steps?
 1) A=xy
 2) 1500 = 3x+y then set equal to y
 3) plug y= 3x+1500 into A=x*y
 4) find the derivative of A(x)
 5) set A'(x) equal to zero and solve this is the length
 6) find the second dervative
 7) answer must be negative since its a max
 8) find the width plug in x into
 original function y= 1500 3x
 multiply both
 8) find the width

postal regulations girth plus length can be no more than 104 inches find the maximum volume and dimensions for this package
 1) 104= 4x + y
 2) V= lwh or xxy or x^2y
 3) sub y= 4x+104 into x^2y
 4) find the derivative
 5) set the derivative V'(x) equal to zero and solve
 6) test with the second derivative answer should be negative since max
 7) plug in the length & width (x) into the original function 4x+104 to get (y) the height
 8) multiply XXY to get the volume

a farmer has 384 feet of fencing for 2adjacent regtangular pens. what is the max area?
 1) 3x+2y = 384
 2) Area = xy
 3) plug in y= 3x+192 into xy
 4) find the derivative of A(x)
 5) set A'(x) equal to zero and solve
 6) use second derivative test to check for max
 7) plug in x into the original function
 3/2x+192 which will give you y
 8) multiply xy for area

population in beginning was 2mil after 3 years it inc to 5 mil what will be the pop in 9 years with exponential growth
 1) exponential growth forumula
 Q(t)= Qoe^kt
 2) 5mil = 2 mil e^k(3)
 3) divid both sides by 2
 4) put ln on both sides to get rid of e which leaves you with
 ln2.5= 3k
 5) divide ln2.5 by 3 to get k
 6) plug k into the function 2e^k(9)
 (side note multiply power k(9) then plug into e then multiply times 2)

solve the initial value problem
f(x)= 3x^2 + 6x 1
f(1)= 6
 1) find the antiderivative of the function
 2) plug in 1 for y and 6 for x
 3) solve for c

x^2
x^3  4
find the indefinate integral
 1) let u = x^3  4
 2) find the derivative of the right side
 3) gives you du = 3x^2
 4) you only need x squared so divide both sides by 3x
 5) then substitute 1/3 * 1/u
 6) substitute the bottom of the function into the integral
 7) gives you 1/3 ln(x^32) + c

find the average value of f(x)= 18x^2+3
over the interval [2, 1]
 1)
 1
 ____ ( integral of f(x))
 1 2
2) 1 [ f(1)  f(2)]

