Question 90369
1.

First draw the picture. In this drawing the thicker line at the top denotes the neighbor's side (which is denoted as N)


{{{drawing( 500, 500, -5, 5, -5, 5,
         locate(0,2.5,N),
         locate(0,-2.5,W),
         blue(line(-3,1.98,3,1.98)),
         blue(line(-3,1.99,3,1.99)),
         blue(line(-3,2,3,2)),
         blue(line(-3,2.01,3,2.01)),
         blue(line(-3,2.02,3,2.02)),

         locate(-3.5,0,L),

         locate(3.5,0,L),

         red(line(3,2,3,-2)),
         blue(line(3,-2,-3,-2)),
         red(line(-3,-2,-3,2))
)}}}


Notice the length of W is equal to N


So the perimeter of this rectangle is 


{{{P=N+W+2L}}}


And the cost function is


{{{C=2N+1.2W+1.2*2L}}}


Now since N=W, we can replace N with W


{{{C=2W+1.2W+1.2*2L}}} replace N with W


{{{C=3.2W+1.2*2L}}} Combine like terms


{{{C=3.2W+2.4L}}} Multiply 1.2 and 2L to get 2.4L




Also, since you have at most is $200, this means the cost of the fence <b>must</b> be less than $200. So we have this inequality


{{{C<200}}}


Now replace C with {{{3.2W+2.4L}}}



{{{3.2W+2.4L<200}}}



Now lets solve for L




{{{2.4L<200-3.4W}}} Subtract 3.4W from both sides


{{{L<(200-3.4W)/2.4}}} Divide both sides by 2.4


{{{L<83.33-1.41667W}}} Divide 


Now since we don't know the value of W, we cannot solve for L. Since there are two unknown variables, there are an infinite number of possible rectangles. For instance, the neighbor's side might be 10 feet


{{{L<83.33-1.41667(10)}}} plug in w=10


{{{L<69.1633}}} evaluate


So in this case there are 69 possible rectangles (if the length is a whole number)

or 

the neighbor's side might be 20 feet


{{{L<83.33-1.41667(20)}}} plug in w=20


{{{L<54.9966}}} evaluate


So in this case there are 53 possible rectangles (if the length is a whole number)

If the length can be any decimal number, then there are an infinite number of rectangles

This list goes on...


So this means we cannot figure out how many possible combinations of rectangles we can make. So I would double check your problem. If that is the correct info, then there's a strong chance that there are an infinite number of rectangles.



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2. 


We previously solved for the cost function


{{{C=3.2W+2.4L}}}


Just replace W and L with x and y


{{{C=3.2x+2.4y}}}



Now lets solve for y


{{{C=3.2x+2.4y}}} Start with the cost function


{{{C-3.2x=2.4y}}} Subtract 3.2x from both sides


{{{(C-3.2x)/2.4=y}}} Divide both sides by 2.4


So we get


{{{y=(C-3.2x)/2.4}}}


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3.

Area:
{{{A=xy}}}



Using our previous answer, plug in {{{y=(C-3.2x)/2.4}}} into {{{A=xy}}}

{{{A=x((C-3.2x)/2.4)}}}



{{{A=(Cx-3.2x^2)/2.4}}} Distribute



Since the cost must be what we have in our pocket, the cost is 200. So c=200


{{{A=(200x-3.2x^2)/2.4}}} Plug in c=200


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4.


If we plot {{{A=(200x-3.2x^2)/2.4}}} as a function of x, we get


{{{ graph( 500, 500, -10, 100, -10, 2000, (200x-3.2x^2)/2.4) }}}


Notice how the graph peaks at a certain value. At this x-value, this x-value is one dimension which forms the maximum area. 


So using a calculator, we get x=41.667

This means that the maximum area is formed when one dimension is 41.667


Now plug in x=41.667 into  {{{y=(200-3.2x)/2.4}}} to find the other dimension



{{{y=(200-3.2(41.667))/2.4}}}


And with a calculator we get


{{{y=27.777}}}



So the maximum area is formed when {{{x=41.667}}} and {{{y=27.777}}}


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5.

Now plug in {{{x=41.667}}} and {{{y=27.777}}} into {{{A=xy}}}


{{{A=(41.667)(27.777)}}}



{{{A=1157.384259}}}


So the maximum area is 1,157.384259 square feet