Where is the largest area, and why is the top flat?

1. 1. Same fence, same area?

   You have $40$ m of fence for a rectangular pen against a long wall, which makes the fourth side. The pen runs $x$ m along the wall and its ends are $y = (40 - x)/2$ m deep. Drag its far corner: the graph below draws its area, $A = xy$.

   Question: As $x$ runs from $0$ to $40$, what does the area do?

   It rises, then falls. The same $40$ m of fence holds nothing at $x = 0$, a pen with no width, and nothing at $x = 40$, a pen with no depth. In between is a best pen, at the top of the graph.

   The figure: A plan, to scale, of a pen against a wall: 8.0 m along the wall and 16.0 m deep, fenced on three sides with 40 m of fence. The wall runs the width of the figure. The area A(x) = x(40 − x)/2 against x. At x = 8.0 it is 128.0 m², and the tangent there has slope 12.0.

2. 2. What a metre trades

   Lengthen the pen by a metre along the wall. It gains a strip down the side, about $y$ square metres, but its ends shorten by half a metre, so it loses a strip along the front, $x/2$ square metres. While the gain beats the loss, the area grows.

   Drag the corner until a metre along the wall gains as much as it loses.

   At $x = 20$, $y = 10$: each metre gains $10$ m² and loses $10$ m², so the slope $A'(x) = y - x/2$ is $0$. That is the best pen, $200$ m², twice as long as it is deep.

   $$A'(x) = y - \tfrac{x}{2} = 20 - x = 0 \quad\text{at}\quad x = 20$$$A^{\prime }(x)$ $=y-\frac{x}{2}$ $=20-x$ $=0$$\text{at}$$x$ $=20$

   The figure: A plan, to scale, of a pen against a wall: 20.0 m along the wall and 10.0 m deep, fenced on three sides with 40 m of fence. The wall runs the width of the figure. Nudged 1.0 m along the wall, it gains a strip of 9.5 m² down the side and loses a strip of 10.0 m² along the front. The area A(x) = x(40 − x)/2 against x. At x = 20.0 it is 200.0 m², and the tangent there has slope 0.0.

3. 3. Why the top is flat

   At the top the gain and the loss balance, so a small change of shape barely changes the area. Here the best pen is nudged $\Delta x = 2$ m along the wall, and its ends shorten by $1$ m.

   Question: How much area does that nudge lose?

   $2$ m². The side strip is $2 \times 9 = 18$ m² and the front strip $20 \times 1 = 20$ m²: only the corner the side strip misses is lost, $(\Delta x)^2/2$. Halve the nudge and the loss falls to a quarter, $0.5$ m². Away from the top it is not small: from $x = 8$ the same nudge gains $22$ m².

   $$A(20 + \Delta x) = 200 - \tfrac12 (\Delta x)^2$$$A(20+\Delta x)$ $=200-\frac{1}{2}(\Delta x{)}^2$

   The figure: A plan, to scale, of a pen against a wall: 20.0 m along the wall and 10.0 m deep, fenced on three sides with 40 m of fence. The wall runs the width of the figure. Nudged 2.0 m along the wall, it gains a strip of 18.0 m² down the side and loses a strip of 20.0 m² along the front. The area A(x) = x(40 − x)/2 against x. At x = 20.0 it is 200.0 m², and the tangent there has slope 0.0.

4. 4. A short wall

   Now the wall is only $14$ m long and the pen must lie along it, so $0 \le x \le 14$. Solving $A'(x) = 0$ still gives $x = 20$: the amber pen, which runs past the end of the wall. Taking it without checking the domain is the slip.

   Question: Which pen holds the most now?

   $x = 14$: a pen $14$ by $13$, $182$ m². On $[0, 14]$, $A'(x) = 20 - x$ stays positive: at the end of the wall a metre more would still gain $13$ m² and lose only $7$. The critical point $x = 20$ is outside the domain, so it is no candidate.

   The figure: A plan, to scale, of a pen against a wall: 14.0 m along the wall and 13.0 m deep, fenced on three sides with 40 m of fence. The wall is 14.0 m long. In amber, the pen where A′ = 0, 20 m by 10 m, which runs past the end of the wall. The area A(x) = x(40 − x)/2 against x, solid on the wall's 14.0 m and dashed beyond it. At x = 14.0 it is 182.0 m², and the tangent there has slope 6.0. In amber, the top of the curve at (20, 200), outside the domain.

5. 5. Check the ends

   So the largest area is at a critical point inside the domain or at one of its ends. The pen drawn now is the best one for its wall. Drag the end of the wall.

   Lengthen the wall until the best pen no longer uses all of it.

   From $20$ m on, the critical point fits on the wall, and the best pen is $20$ by $10$, $200$ m², however long the wall. Shorter than $20$ m, $A'$ stays positive to the end and the best pen takes the whole wall.

   $$\max_{0 \le x \le W} A(x) = \text{the largest of } A(0),\ A(W) \text{ and } A(c) \text{ where } A'(c) = 0,\ 0 < c < W$$$\underset{0\le x\le W}{\max }A(x)$ $=\text{the largest of }A(0),A(W)\text{ and }A(c)\text{ where }{A}^{\prime }(c)$ $=0,0$ $<c$ $<W$

   The figure: A plan, to scale, of a pen against a wall: 20.0 m along the wall and 10.0 m deep, fenced on three sides with 40 m of fence. The wall is 26.0 m long. The area A(x) = x(40 − x)/2 against x, solid on the wall's 26.0 m and dashed beyond it. At x = 20.0 it is 200.0 m², and the tangent there has slope 0.0.

The film, as said, 1:46

1. 0:00

   $40$ m of fence makes three sides of a pen against a wall: $x$ m along it, ends $y = (40 - x)/2$ m deep.

2. 0:07

   Run $x$ from short to long. The area rises to a top, then falls: the same fence does not hold the same area.

3. 0:19

   Add a metre along the wall. The pen gains a strip down the side, about $y$ m², and loses one along the front, $x/2$ m².

4. 0:30

   At $x = 8$ a metre gains $16$ m² and loses $4$: the area is still growing.

5. 0:35

   At $x = 20$ a metre gains $10$ and loses $10$, so $A' = 0$: the best pen, $20$ by $10$, $200$ m².

6. 0:44

   There a nudge barely matters. Lengthen by $2$ m: the side strip is $18$ m², the front strip $20$, a loss of only $2$.

7. 0:54

   Halve the nudge and the loss falls to a quarter, $0.5$ m². That is why the top of the graph is flat.

8. 1:02

   Now the wall is only $14$ m long. The pen must lie along it, so $x$ is at most $14$.

9. 1:12

   The slip solves $A'(x) = 0$ anyway and reports $x = 20$: the amber pen, which runs past the end of the wall.

10. 1:18

    On the wall's $14$ m the area only rises, so the best pen uses all of it: $14$ by $13$, $182$ m².

11. 1:28

    Lengthen the wall and the best pen grows with it, up to $20$ m. From there it stays $20$ by $10$.

12. 1:39

    The largest value is at a critical point inside the domain or at an end: find every candidate, and compare.