Single-Variable-Calculus · Unit 10 · Video 1 · Interactive Practice

Two Squares From One Wire: The Critical Point Gives the Least Area

IKey Formulas

Formula	Name	What it tells you
$A(x) = \left(\tfrac{x}{4}\right)^2 + \left(\tfrac{1-x}{4}\right)^2,\ \ 0 < x < 1$	Total area	Pieces $x$ and $1 - x$ bent into squares of sides $\tfrac{x}{4}$ and $\tfrac{1-x}{4}$
$A'(x) = \tfrac{x}{8} - \tfrac{1-x}{8}$	Derivative (chain rule on the second term)	$A' = 0$ only at $x = \tfrac12$ : one critical point, critical value $A(\tfrac12) = \tfrac{1}{32}$
$\displaystyle\lim_{x\to 0^+} A = \lim_{x\to 1^-} A = \tfrac{1}{16}$	Ends by limits	The cut can't sit at $0$ or $1$ , so the largest area is approached, never attained
$A''(x) = \tfrac18 + \tfrac18 = \tfrac14 > 0$	Second-derivative check	Concave up everywhere, so $x = \tfrac12$ is a minimum

Key Insight: The minimum value is $\tfrac{1}{32}$ ; it is achieved at $x = \tfrac12$ . Setting $A' = 0$ answered the question "where is the least area?", the opposite of what was asked.

IIVisualization 1 — Where Should the Cut Go?

Every cut $0 < x < 1$ gives two squares and one point on the graph of $A(x)$ .

IIIVisualization 2 — From $A(x)$ to the Answer

The one critical point, checked against both ends, turns out to be the least area, not the largest.

Step 1

Step 1 — Area as a function of $x$

A(x) = \left(\frac{x}{4}\right)^2 + \left(\frac{1-x}{4}\right)^2 = \frac{x^2}{16} + \frac{(1-x)^2}{16}

Step 2 — Differentiate

\left(\frac{x^2}{16}\right)' = \frac{x}{8}, \qquad \left(\frac{(1-x)^2}{16}\right)' = \frac{2(1-x)(-1)}{16} = -\frac{1-x}{8}

A'(x) = \frac{x}{8} - \frac{1-x}{8}

Step 3 — Critical point and critical value

A' = 0 \iff x = 1 - x \iff x = \tfrac12

A\left(\tfrac12\right) = \left(\tfrac18\right)^2 + \left(\tfrac18\right)^2 = \tfrac{1}{32}

Step 4 — Ends, by limits (cut at $0$ or $1$ leaves one piece)

\lim_{x\to 0^+} A = 0 + \left(\tfrac14\right)^2 = \tfrac{1}{16}, \qquad \lim_{x\to 1^-} A = \left(\tfrac14\right)^2 + 0 = \tfrac{1}{16}

Step 5 — Compare, and confirm with $A''$

\tfrac{1}{32} = \tfrac12 \cdot \tfrac{1}{16}, \qquad A''(x) = \tfrac18 + \tfrac18 = \tfrac14 > 0

Minimum value

\tfrac{1}{32}

, achieved at

x = \tfrac12

; the largest area

\tfrac{1}{16}

is only a limit.

IVVisualization 3 — Approached, Never Reached

As one square shrinks to nothing, $A(x) = \tfrac{1}{16} - \tfrac{x(1-x)}{8}$ climbs toward $\tfrac{1}{16}$ but never gets there.

shrink x = 0.250

💡 The question asked for a largest area that does not exist under its own rules; the most informative answer is $\tfrac{1}{16}$ , approached as one square vanishes.

VQuiz Questions

Problem 1 · Area at a Given Cut

Given: A wire of length $1$ is cut at $x = \tfrac14$ , and each piece is bent into a square — find the total area enclosed.

✅ Correct! Sides

\tfrac{1}{16}

and

\tfrac{3}{16}

give

\tfrac{1}{256} + \tfrac{9}{256} = \tfrac{5}{128}

❌ Not quite. That squares the piece lengths themselves. A piece of length

x

makes a square of side

\tfrac{x}{4}

, since the perimeter is

4 \times

side.

❌ Close, but… you divided

x^2 + (1-x)^2

4

. The side is

\tfrac{x}{4}

, so the area is

\tfrac{x^2}{16}

: the

4

gets squared too.

❌ Not quite.

\tfrac{1}{32}

is the area at the middle cut

x = \tfrac12

. At

x = \tfrac14

the squares are unequal, and the area is larger than

\tfrac{1}{32}

❌ Not quite. Find each square's side first: piece length divided by

4

Show solution

The pieces have lengths $\tfrac14$ and $\tfrac34$ , so the squares have sides $\tfrac{1}{16}$ and $\tfrac{3}{16}$ :

A\left(\tfrac14\right) = \left(\tfrac{1}{16}\right)^2 + \left(\tfrac{3}{16}\right)^2 = \frac{1}{256} + \frac{9}{256} = \frac{10}{256} = \frac{5}{128}

Check against the bounds: $\tfrac{1}{32} = \tfrac{4}{128} < \tfrac{5}{128} < \tfrac{8}{128} = \tfrac{1}{16}$ .

Problem 2 · The Critical-Point Trap

Given: A student solves $A'(x) = \tfrac{x}{8} - \tfrac{1-x}{8} = 0$ , gets $x = \tfrac12$ , and reports "the largest area is $\tfrac{1}{32}$ " — which statement correctly answers "find the largest area" for $0 < x < 1$ ?

✅ Correct! The ends give

\tfrac{1}{16}

only as limits, and

A(x) < \tfrac{1}{16}

for every cut, so the largest value is approached but never reached.

❌ That is the trap.

\tfrac{1}{32}

is the least area. Comparing with the ends, where

A \to \tfrac{1}{16}

, shows the critical point is a minimum.

❌ Close, but… a cut at

x = 0

leaves only one piece, so

x = 0

is excluded. The value

\tfrac{1}{16}

is a limit as

x \to 0^+

, not a value of

A

❌ Not quite.

\tfrac18

is the side of each square at

x = \tfrac12

, not an area; and

x = \tfrac12

gives the least area anyway.

Show solution

Check all candidates. Critical point: $A(\tfrac12) = \tfrac{1}{32}$ . Ends (excluded, so use limits):

\lim_{x\to 0^+} A(x) = 0 + \left(\tfrac14\right)^2 = \tfrac{1}{16}, \qquad \lim_{x\to 1^-} A(x) = \tfrac{1}{16}.

Also $\tfrac{1}{16} - A(x) = \tfrac{x(1-x)}{8} > 0$ on $0 < x < 1$ , so $A$ never equals $\tfrac{1}{16}$ . The minimum value is $\tfrac{1}{32}$ at $x = \tfrac12$ ; there is no maximum, only the limit $\tfrac{1}{16}$ as one square shrinks to nothing.

Problem 3 · Minimum Value vs. Where It Is Achieved

Given: A wire of length $12$ is cut into pieces $x$ and $12 - x$ , each bent into a square, so $A(x) = \left(\tfrac{x}{4}\right)^2 + \left(\tfrac{12-x}{4}\right)^2$ on $0 < x < 12$ — find the minimum value of $A$ and where it is achieved.

What is the minimum value?

Where is the minimum achieved?

✅ Correct! The minimum value is

\tfrac92

, a value of

A

; it is achieved at

x = 6

, a place on the wire. Two questions, two answers.

❌ That answers the other question.

6

is where the minimum occurs (the critical point). The minimum value is

A(6)

❌ Close, but…

\left(\tfrac32\right)^2 = \tfrac94

is the area of one square. At

x = 6

there are two equal squares.

❌ Not quite.

9

is the limit at the ends, where one square of side

3

remains: the largest area, approached but not attained.

❌ That answers the other question.

\tfrac92

is the minimum value, an area. Where it is achieved is a cut position

x

❌ Not quite. Solve

A'(x) = \tfrac{x}{8} - \tfrac{12-x}{8} = 0

: that gives

x = 12 - x

❌ Not quite.

x = 0

is excluded, and the ends are where the area is largest.

❌ Check the value. Evaluate

A

at the critical point.

❌ Check the location. Set

A'(x) = 0

Show solution

A'(x) = \frac{2x}{16} + \frac{2(12-x)(-1)}{16} = \frac{x}{8} - \frac{12-x}{8} = 0 \iff x = 12 - x \iff x = 6

Each square then has side $\tfrac64 = \tfrac32$ , so

A(6) = \left(\tfrac32\right)^2 + \left(\tfrac32\right)^2 = \tfrac94 + \tfrac94 = \tfrac92.

$A'' = \tfrac18 + \tfrac18 = \tfrac14 > 0$ , so this is a minimum; the ends give $\lim A = 3^2 = 9$ . Minimum value: $\tfrac92$ . Where achieved: $x = 6$ . Saying "the minimum is $6$ " confuses the two.

Problem 4 · Transfer: A Square and a Triangle

Given: A wire of length $1$ is cut; the piece of length $x$ is bent into a square and the piece $1 - x$ into an equilateral triangle, so $A(x) = \dfrac{x^2}{16} + \dfrac{\sqrt3\,(1-x)^2}{36}$ on $0 < x < 1$ — find where the minimum is achieved and the value the area approaches at the better end.

Where is the minimum achieved?

What does the area approach at the better end?

✅ Correct! The single critical point

x = \tfrac{4\sqrt3}{9+4\sqrt3} \approx 0.435

is again a minimum; the largest area,

\tfrac{1}{16} = 0.0625

, is approached as the whole wire goes into the square.

❌ Not quite. The middle cut worked for two squares because both terms had the same coefficient. Here they don't: solve

\tfrac{x}{8} = \tfrac{\sqrt3\,(1-x)}{18}

❌ Close, but…

\tfrac{9}{9+4\sqrt3}

is the length of the triangle's piece,

1 - x

. The question asks for

x

, the square's piece.

❌ Not quite.

\tfrac{9}{4\sqrt3} \approx 1.30

is outside

0 < x < 1

. Collect the

x

terms:

9x = 4\sqrt3\,(1 - x)

❌ That is the other end. As

x \to 0^+

(all triangle)

A \to \tfrac{\sqrt3}{36} \approx 0.0481

; as

x \to 1^-

(all square)

A \to \tfrac{1}{16} = 0.0625

, which is larger.

❌ Not quite. At each end one shape vanishes, so only one term survives the limit.

❌ Close, but…

\tfrac{\sqrt3}{4}

is the area of an equilateral triangle of side

1

. A wire of length

1

makes a triangle of side

\tfrac13

, area

\tfrac{\sqrt3}{36}

, and that is the smaller end anyway.

❌ Check the location. Set

A'(x) = \tfrac{x}{8} - \tfrac{\sqrt3\,(1-x)}{18} = 0

❌ Check the ends. Take both limits and compare.

Show solution

An equilateral triangle of perimeter $1 - x$ has side $\tfrac{1-x}{3}$ and area $\tfrac{\sqrt3}{4}\left(\tfrac{1-x}{3}\right)^2 = \tfrac{\sqrt3\,(1-x)^2}{36}$ . Differentiate:

A'(x) = \frac{x}{8} - \frac{\sqrt3\,(1-x)}{18} = 0 \iff 9x = 4\sqrt3\,(1-x) \iff x = \frac{4\sqrt3}{9+4\sqrt3} \approx 0.435

$A''(x) = \tfrac18 + \tfrac{\sqrt3}{18} > 0$ , so this is a minimum, with value $\approx 0.0272$ . Ends by limits:

\lim_{x\to 0^+} A = \frac{\sqrt3}{36} \approx 0.0481, \qquad \lim_{x\to 1^-} A = \frac{1}{16} = 0.0625.

The larger, $\tfrac{1}{16}$ , is the largest area, approached (not attained) as the triangle's piece shrinks to nothing.

Solved: 0 / 4

Two Squares From One Wire: The Critical Point Gives the Least Area

IKey Formulas

IIVisualization 1 — Where Should the Cut Go?

IIIVisualization 2 — From A(x)A(x)A(x) to the Answer

IVVisualization 3 — Approached, Never Reached

VQuiz Questions

IIIVisualization 2 — From $A(x)$ to the Answer