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With a $40\mathrm{m}$ long fence, it is desired to create a trapezoidal region with a base wall. What is the largest area that can be created? How can I calculate this question?

I've tried all $3$ sides of trapezoid $\frac{40}3$ and creating $3$ triangles. And that : For only $3$ sides from $40\mathrm{m}$:

Let the top side opposite the wall be $x\mathrm{m}$. The other two sides are therefore $\frac{40 - x}2$. The area, $A$, of the trapezoid is therefore $x\frac{40 - x}2 = 20x - \frac{x^2}2$. After that derivatives and etc... But i'm not sure. (isosceles trapezoid)

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  • $\begingroup$ What have you tried? Perhaps start by labeling unknowns and writing the relationships between them. $\endgroup$ – Andrew Chin Jun 22 '20 at 18:32
  • $\begingroup$ @AndrewChin I've tried all 3 sides of trapezoid 40/3 and creating 3 triangles. And that : For only 3 sides from 40 metres: Let the top side opposite the wall be x metres. The other two sides are therefore (40 - x)/2. The area, A, of the trapezoid is therefore x(40 - x)/2 = 20x - x^2/2. After that derivatives and etc... But i'm not sure. $\endgroup$ – Kaan Deniz Jun 22 '20 at 18:41
  • $\begingroup$ This should go in your post when asking the question. $\endgroup$ – Andrew Chin Jun 22 '20 at 18:41
  • $\begingroup$ I'm new here. Sorry about that. $\endgroup$ – Kaan Deniz Jun 22 '20 at 18:42
  • $\begingroup$ Are you assuming the trapezoid is isosceles? $\endgroup$ – ir7 Jun 22 '20 at 18:46
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Knowing that the region to be enclosed must be a trapezoid, this means the middle segment must be parallel to the existing wall. Since the formula for the area of a trapezoid is $$A = \frac{h}{2}(b_1 + b_2),$$ this means that if the bases are constant and the height is constant, the area is the same even if the location of the two bases are are "shifted" relative to each other. So for a trapezoid of given fixed area, the one that minimizes the perimeter is the one that has the lateral sides, which we will call $l_1$ and $l_2$, equal--that is to say, the trapezoid is isosceles. When we reverse this reasoning, it follows that for a fixed length of fencing, we only ever need to consider dividing the fencing into three lengths, say $x, y, z$, such that $x = z$, since any other choice with $x \ne z$ yields a trapezoid with inferior area. Since $x + y + z = 40$, this gives us a second constraint, so that there is one free variable to consider, say $y$, that uniquely determines how the fence is cut.

However, for a given partitioning of the fencing, there are numerous ways to place it against the wall to enclose a trapezoidal area. At one extreme, we can simply lay it flat against the wall and enclose no area. At the other, we can put the ends of the fence as close to each other as possible along the wall, which, depending on whether $2x < y$, might enclose no area, or a triangular area. This suggests letting the angle $\theta$ be the internal angle between the wall and one of the lateral sides. We have the base $b_2 = y$, the two lateral sides $x = (40 - y)/2$, and we need to solve for the height $h$ and wall-side base $b_1$ for a given angle $\theta$.

Trigonometry gives us $$\sin \theta = \frac{h}{x}, \\ b_1 = b_2 + 2 x \cos \theta.$$ So the area as a function of $y$ and $\theta$ is $$A(y, \theta) = \frac{40-y}{2} \sin \theta \left(y + \frac{40-y}{2} \cos \theta\right).$$ Now I leave the rest as an exercise to compute the values of $y, \theta$ such that $A$ is maximized.

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