Maximum area inside a fenced garden - - - my challenge question

[lookagain]

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Suppose you have a garden against a straight wall (using the wall as one side).

You must use 120 feet of fencing to form three straight sides and use them and
the wall in such a way so as to make an isosceles trapezoidal shape to . . . < ---- edit
enclose the garden.


What is the maximum area in square feet possible for the garden?



. . . . . . . ___________________
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. . . . . . / . . . . . garden . . . . . .\
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. . . . . . . wall side of the garden

 

[daon2]

I made some assumptions, since I do not know what an equilateral trapezoid is. I assumed the (possiblly) non-parallel sides were equal.

Let \(\displaystyle \theta\) be the angle, between 0 and pi, that the sides make with the wall. Let y we the side of the garden parallel to the wall, W the length of the wall, and L the lengths of the congruent sides. Then the height, \(\displaystyle H=L\sin\theta\).

Maximize, with respect to \(\displaystyle H,y\):
\(\displaystyle A= \frac{H}{2}(y+W) = \frac{1}{2}(Hy) + \frac{W}{2}H\)

Given constraint: \(\displaystyle 120=2L+y+W\implies \frac{\sin\theta}{2}(120-y-W)=H\)

\(\displaystyle A_y = \frac{1}{2}(y+W)H_y + \frac{1}{2}H = \frac{1}{2}H-\frac{H}{4L}(y+W)\)

Setting this to 0 gives: \(\displaystyle y=2L-W\)

To which the constraint assigns: \(\displaystyle L=30\), and so \(\displaystyle y=60-W\)

So \(\displaystyle A_{max} = HL = L^2\sin\theta = 900\sin\theta\)

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Done another way (using rectangle+two triangles), we have \(\displaystyle A = (W-2x)H + 2(\frac{1}{2}xH) = (W-x)H\) where \(\displaystyle x=L\cos\theta\) and the new constraint given by: \(\displaystyle 120=2(W+\frac{H}{\sin\theta}-x)\implies H=(60-W+x)\sin\theta\).

\(\displaystyle A_x = H_x(W-x)-H = (W-x)\sin\theta-(60-W+x)\sin\theta\)

Setting this to 0 yields:

\(\displaystyle x=W-30\), to which the constraint assigns \(\displaystyle H = 30\sin\theta\), and so \(\displaystyle x=30\cos\theta\)

Plugging this into \(\displaystyle A\) we get: \(\displaystyle (30)30\sin\theta=900\sin\theta\)

edit: Of couse, \(\displaystyle \sin\theta = \frac{H}{L} = \sqrt{L^2-x^2}/L\), which is greatest at theta=pi/2. So the trapezoid must be a rectangle

 

[lookagain]

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Plugging this into A get: \(\displaystyle (30)30\sin\theta=900\sin\theta\)

edit: Of couse, \(\displaystyle \sin\theta = \frac{H}{L} = \sqrt{L^2-x^2}/L\), which is greatest at theta=pi/2. So the trapezoid must be a rectangle

Are you stating that A (the number of square feet) is a maximum when theta = pi/2?

And if so, then \(\displaystyle 900\sin(\theta) \ = \ 900\sin (pi/2) \ = \ 900(1) \ = \ 900?\)



What about if the sides of the fence were 30 ft., 60 ft., and 30 ft.** (against the wall)?

If those were formed to make the shape of a rectangular garden, then A would equal 1800 sq. ft.



Or, what if the 30 ft. sides were allowed to not be perpendicular to the 60 ft. side?


Or, what if the sides were not the dimensions as in the scenario ** above?

 

[daon2]

You're right, for some reason or another I decided to include W, the length of the wall, in the constraint. I should have realized the answer was odd as W did not show up in the end!

Revised:

\(\displaystyle A= \frac{H}{2}(y+W) = \frac{1}{2}(Hy) + \frac{W}{2}H\)

Given constraint: \(\displaystyle 120=2L+y\implies \frac{\sin\theta}{2}(120-y)=H\)

\(\displaystyle A_y = \frac{1}{2}(y+W)H_y + \frac{1}{2}H = \frac{1}{2}H-\frac{H}{4L}(y+W)\)

Setting this to 0 gives: \(\displaystyle y=2L-W\)

To which the constraint assigns: \(\displaystyle L=\dfrac{120+W}{4}\), and so \(\displaystyle y=60-\dfrac{W}{2}\)

So \(\displaystyle A_{max} = \dfrac{H}{2}\left(60-\dfrac{W}{2}+W\right)= \left(30+\dfrac{W}{4}\right)^2 \sin(\theta)\)

Now, \(\displaystyle \sin\theta\) is quite ugly, but algebraically can be gotten from the fact that \(\displaystyle 2L\cos(\theta) + y = W\), which will show that \(\displaystyle \sin\theta\) is equal to \(\displaystyle \sqrt{1-\left(\dfrac{3W-120}{120+W}\right)^2} = \sqrt{8}\dfrac{\sqrt{120W-W^2}}{120+W}\). This at least makes sense, as the area should be 0 when \(\displaystyle W=0, W=120\)

 

[lookagain]

A hexagon with a set perimeter, P, has its maximum area when it is a regular hexagon.

Each side would have P/6 units in length.

The wall's edge is acting as a long diagonal through a hexagon, forming an isosceles trapezoid.


This trapezoid would have a maximum area when it is half of a regular hexagon, the wall
being the long diagonal of the regular hexagon.


Then each side of the trapezoidal shape would be 40 feet to maximize the area of the garden,
as 120 ft./3 = 40 ft.


This trapezoid will be composed of three (non-overlapping) equilateral triangles
with side lengths of 40 feet.


The area of an equilateral triangle, with side length s, \(\displaystyle \ is \ \ \dfrac{s^2\sqrt{3}}{4}.\)


Then the total maximum area of the garden is \(\displaystyle \ 3\bigg(\dfrac{(40 \ ft.)^2\sqrt{3}}{4}\bigg) \ =\)


\(\displaystyle 1200\sqrt{3} \ \ square \ feet\)