Calculus Max Min Area Word Problems
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In the Area
Max Min Area Word Problems

 

 So we spent all that time with maxima and minima

 because most of the problems you will see deal with finding 
 the maximum or minimum amount of something with a given amount of input.
 
 That is actually a good definition of what engineering is all about.
 Things like:
 How big a bridge can you build with a certain amount of materials.
 How many people can you push through the Haunted Mansion ride in a day.
 What's the cheapest computer that can fly a plane safely.
 How many cows can you keep alive on a certain farm 
    (Note: this is not the exact same problem as #2 above).
 
 Anyway, the drill is we will be given some problem, generally as a paragraph.
 We convert that to a function
 (like it was in Algebra, this is usually the hardest part).
 And use the maxima and minima techniques from the last few chapters
 to find the answer.
 
 Area problems are some of the least complex types so we'll start with those.
 There are two types of area problems. Fence types and surface types.
 Generally fence types deal with finding 2 dimensional areas
 and surface types deal with finding 3 dimensional volumes.
 
 In the fence type, we usually need to enclose the maximum area possible
 with a certain amount of fence, or we might need to enclose a certain area 
 with the minimum amount of fence.

 

 This will give us these two equations:
 

 2W + 2L = Total Length of Fence

 Area Fenced = W x L

 
 Sometimes we are given the total amount of fence
 and want to know how much area can be fenced.
 
 We are going to do the maxima and minima stuff with our equations above, 
 but they have too many variables.
 We can cut that down by solving the first equation for L ...
 
2W + 2L =   Total Length of Fence
2L =   Total Length of Fence - 2W
 
L =   Total Length of Fence  - W

2
 
 Remember, in this style of problem the total length of fence is given,
 it's a number so that messy fraction in the middle will just be some number.
 
 So now we just substitute this value in for L in the second equation and get:
 
Area Fenced =   W x L
 
Area Fenced =  

W x (

Total Length of Fence  - W) 

2

 

 Multiply this out and rearrange a bit, and we get:
 

Area Fenced =  

-W2 + (

Total Length of Fence )W

2
 
 This is actually just an ordinary quadratic function in disguise.
 An ordinary quadratic generally looks like this:
 

F(X) = AX2 + BX + C

 
 Area Fenced is playing the part of F(X)
 W is playing the part of X.
 A is -1
 B is the goofy Total Length of Fence/2 thing
 C is zero, so it doesn't even show up
 
 So take the derivative of this equation, find where the slope is zero
 and that or an end point will be the answer candidates.
 
 There are two common variations of this problem.
 In one variation, one of the sides of the area to be fenced is a river 
 or something else that doesn't need to be fenced
 

 

 Since we get one of the W's for free, that makes the "Length of Fence" equation:
 

 W + 2L = Total Length of Fence

 
 Other than that, it's the same problem.
 
 In the other variation, you have to fence in two areas

 

 Since we need an extra L, that makes the "Length of Fence" equation:
 

 2W + 3L = Total Length of Fence

 
 Other than that, it's the same problem.
 
 Neither of these variations change the area equation.
 It will still be:

 Area Fenced = W x L

 
 Yeesh, enough already, let's do one!
 OK, we have 1200 feet of fence
 and we need to enclose as big an area as we can into two sections...
 

 

 
 Here we go. Since we have two sections and 1200 feet of fence,
 one of our equations will be:
 

 2W + 3L = 1200

 
 The area equation is still:

 Area Fenced = W x L

 
 We want to find the maximum area,
 so we need to solve the first equation for one of the variables
 (either one will do), 
 then we need to substitute that into the area equation, 
 find the first derivative of the area equation,
 and do our maxima stuff.
 We are only looking for the maximum area, so we don't need the minima this time.
 
 OK, lets solve the first equation for W
 
 2W + 3L =   1200
2W =   1200 -  3L
 W =   600 -  3  L

2
 
 So substituting for W in the area equation and multiplying out ...
 
A =   L x W
 A =   L x (600 -  3  L)

2
A =   -3 L2 + 600L

2
 
 A local minima or maxima will happen anywhere the slope is zero,
 so we need the derivative of the area equation, we'll call it A':
 
A =   -3  L2 + 600L

2
A' =  

-3L + 600

 
 Now find the L value where A' = 0 ...
 
A' =   0 = -3L + 600
3L =   600
L =   200
 
 So where the slope is zero, L = 200.
 Now find W at that point.
 
 We know that 2W + 3L = 1200 so:
 

 2W + 3L = 1200

 2W + 3(200) = 1200

 2W + 600 = 1200

 2W = 600
 W = 300
 
 So the first suspect point is W = 300, L = 200
 
 What are the end points?
 Well the lengths can't be negative,
 so the endpoints are where one or the other (L or W) is zero. 
 Find these points ...
 
2W + 3L = 1200  2W + 3L = 1200
 2W + 3(0) = 1200  2(0) + 3L = 1200
2W = 1200 3L = 1200
W = 600 L = 400
 
 So the candidates for the maximum area are:
 
W = 600, L = 0 an endpoint
W = 0, L = 400 the other endpoint
W = 300, L = 200 the A' = 0 point
 
 Solving the area equation for these points:
 
A = W x L A = W x L A = W x L
A = 600 x 0 A = 0 x 400 A = 300 x 200
A = 0 A = 0 A = 60,000
(not very big) (not any better) (A Winner!)
 

   copyright 2005 Bruce Kirkpatrick

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