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jillmray
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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
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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
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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
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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)
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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
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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^3-2) + c
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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)]
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