Single-Variable-Calculus · Unit 10 · Video 2 · Interactive Practice

The Open Box of Fixed Volume: Least Material When x = 2y

IKey Formulas

Formula	Step	Where it comes from
$V = x^2 y$ (fixed), $\quad A = x^2 + 4xy$	Setup	Square base $x \times x$ , height $y$ ; bottom plus four $x \times y$ sides, no top
$A(x) = x^2 + \dfrac{4V}{x}$ , $\quad A'(x) = 2x - \dfrac{4V}{x^2} = 0 \iff x^3 = 2V$	Method 1: substitution	Solve the constraint for $y = V/x^2$ ; $A \to \infty$ as $x \to 0^+$ and as $x \to \infty$ , and $A'' = 2 + \dfrac{8V}{x^3} > 0$
$0 = 2xy + x^2 y' \Rightarrow y' = -\dfrac{2y}{x}$ , $\quad \dfrac{dA}{dx} = 2x + 4y + 4xy' = 2x - 4y$	Method 2: implicit differentiation	Differentiate the constraint and the area with $y = y(x)$ ; $\dfrac{dA}{dx} = 0 \Rightarrow \dfrac{x}{y} = 2$
$x = 2^{1/3}V^{1/3}$ , $\ y = 2^{-2/3}V^{1/3}$ , $\ A_{\min} = 3 \cdot 2^{2/3} V^{2/3}$	The answers	Dimensionless: $\dfrac{x}{y} = 2$ and $\dfrac{A_{\min}}{V^{2/3}} = 3 \cdot 2^{2/3} \approx 4.76$

Key Insight: Only $V$ is fixed, so $x$ runs over $0 < x < \infty$ and the area is infinite at both ends; the single critical point must be the minimum. At every size, the best open box is twice as wide as it is tall.

IIVisualization 1 — Same Volume, Different Boxes

With $V = 1$ fixed, the area $x^2 + 4V/x$ blows up at both ends and bottoms out once, at $x = 2^{1/3}$ .

Base side x 0.80

IIIVisualization 2 — Method 2 on the Constraint Curve

Along $x^2 y = 4$ , implicit differentiation gives $dA/dx = 2x - 4y$ , which vanishes only where $x/y = 2$ .

Base side x 1.20

IVVisualization 3 — One Shape at Every Size

Because $V$ cancels in $x/y$ , every optimal box is $2 : 1$ , while its least area grows like $V^{2/3}$ .

Volume V 4.00

💡 Scaling law: doubling the volume multiplies the least area by $2^{2/3} \approx 1.59$ , not by $2$ ; from $V = 4$ to $V = 32$ the volume grows $8$ -fold and the area only $4$ -fold, from $12$ to $48$ .

VQuiz Questions

Problem 1 · The Critical Point

Given: an open-top box with a square base of side $x$ and height $y$ must hold $V = 32$ cubic inches — find the base side $x$ that uses the least material.

✅ Correct!

x^3 = 2V = 64

gives

x = 4

, and then

y = 32/16 = 2

: a

2 : 1

box.

❌ That is the height. With

x = 4

the height is

y = V/x^2 = 32/16 = 2

; the question asks for the base side.

❌ That is the cube, the answer for a closed box. With no top,

A = x^2 + 4V/x

, and

A' = 0

gives

x^3 = 2V

, not

x^3 = V

❌ The derivative of $4V/x$ is off.

\dfrac{d}{dx}\left(4V x^{-1}\right) = -4V x^{-2} = -\dfrac{4V}{x^2}

, not

-\dfrac{4V}{x}

; the wrong version leads to

x^2 = 2V = 64

❌ Not quite. Substitute

y = V/x^2

into

A = x^2 + 4xy

, differentiate, and set

A'(x) = 0

Show solution

The constraint $V = x^2 y = 32$ gives $y = 32/x^2$ . Substituting into $A = x^2 + 4xy$ :

A(x) = x^2 + 4x \cdot \frac{32}{x^2} = x^2 + \frac{128}{x}.

A'(x) = 2x - \frac{128}{x^2} = 0 \;\Rightarrow\; 2x^3 = 128 \;\Rightarrow\; x^3 = 64 \;\Rightarrow\; x = 4.

Since $A \to \infty$ as $x \to 0^+$ and as $x \to \infty$ , this single critical point is the minimum. The height is $y = 32/16 = 2$ , so $x/y = 2$ , as the general answer $x = 2^{1/3}V^{1/3} = (2 \cdot 32)^{1/3} = 4$ predicts.

Problem 2 · Checking the Ends

Given: $A(x) = x^2 + \dfrac{4V}{x}$ with $V$ fixed and the single critical point $x = 2^{1/3}V^{1/3}$ — decide the range of $x$ and what kind of critical point it is.

What is the range of x?

What is the critical point?

✅ Correct!

x

has no natural upper limit, the area is infinite at both ends, and a graph that comes down from

\infty

and goes back up with one turn must turn at its minimum.

❌ $A$ is not fixed here; only $V$ is. The height

y = V/x^2

can be as small as we like, so

x

can be as large as we like.

❌ $x$ is not limited by the cube. A base wider than

V^{1/3}

just means a height below

V^{1/3}

; the flat box with

x = 10V^{1/3}

is allowed.

❌ $x = 0$ is not a box. A side must have positive length, and

A(x) = x^2 + 4V/x

is undefined at

0

❌ A zero slope can be a minimum just as well. Here the area is infinite at both ends, so the one turning point is the lowest point, not the highest.

❌ The end check already decides it.

A''

is an alternative, not a requirement:

\lim_{x \to 0^+} A = \infty

and

\lim_{x \to \infty} A = \infty

with one critical point between leave only a minimum.

❌ The area does have a least value.

A

is continuous on

0 < x < \infty

and infinite at both ends, so it attains a minimum, at the critical point.

❌ Check the range. Which of

x

and

A

is constrained? Only

V

is fixed.

❌ Check the type. Compute

\lim_{x \to 0^+} A

and

\lim_{x \to \infty} A

Show solution

Range. A side has positive length, so $x > 0$ . Nothing bounds $x$ from above: the height $y = V/x^2$ just shrinks toward $0$ . So $0 < x < \infty$ .

Ends.

\lim_{x \to 0^+}\left(x^2 + \frac{4V}{x}\right) = \infty \quad (\text{the } 4V/x \text{ term, the tall thin box}),

\lim_{x \to \infty}\left(x^2 + \frac{4V}{x}\right) = \infty \quad (\text{the } x^2 \text{ term, the flat box}).

The graph comes down from $\infty$ , turns once, and goes back up, so the single critical point $x = 2^{1/3}V^{1/3}$ is the minimum. The second derivative agrees: $A''(x) = 2 + \dfrac{8V}{x^3} > 0$ for $x > 0$ .

Problem 3 · The Full Answer for V = 500

Given: an open-top box with a square base must hold $V = 500$ cubic inches — find the height of the best box and its least surface area.

What is the height y?

What is the least area?

✅ Correct!

x = 10

y = 5

, and

A = 100 + 200 = 300

; check:

3 \cdot 2^{2/3} \cdot 500^{2/3} = 3 \cdot 1000^{2/3} = 300

❌ That is the cube's height. The base side is

x = (2 \cdot 500)^{1/3} = 10

, and the height is

y = 500/10^2 = 5

, half the base side.

❌ The ratio is upside down. The base side is twice the height,

x/y = 2

, not

y/x = 2

; and

10 \cdot 10 \cdot 20 = 2000

, not

500

❌ That is only the four sides.

4V/x = 2000/10 = 200

; add the bottom,

x^2 = 100

❌ That counts a top. The box is open: the area is the bottom

x^2 = 100

plus the sides

4xy = 200

, with no second

x^2

❌ That is only the bottom.

x^2 = 100

; the four sides add

4xy = 4 \cdot 10 \cdot 5 = 200

❌ Check the height. First

x^3 = 2V

, then

y = V/x^2

❌ Check the area. Evaluate

A = x^2 + 4xy

at the optimal

x

and

y

; the sides are twice the bottom.

Show solution

Base side: $x^3 = 2V = 1000$ , so $x = 10$ in.

Height:

y = \frac{V}{x^2} = \frac{500}{100} = 5 \text{ in}, \qquad \frac{x}{y} = \frac{10}{5} = 2.

Least area: when $x^3 = 2V$ we have $4V = 2x^3$ , so the sides $\dfrac{4V}{x} = 2x^2$ are twice the bottom:

A = x^2 + 4xy = 100 + 4 \cdot 10 \cdot 5 = 100 + 200 = 300 \text{ in}^2.

The formula agrees: $A_{\min} = 3 \cdot 2^{2/3} V^{2/3} = 3\,(2V)^{2/3} = 3 \cdot 1000^{2/3} = 3 \cdot 100 = 300$ .

Problem 4 · A Costlier Bottom (Implicit Differentiation)

Given: an open-top box with square base $x$ and height $y$ holds a fixed volume $V = x^2 y$ , but the bottom costs $3$ times as much per square inch as the sides, so the cost is proportional to $C = 3x^2 + 4xy$ — find the shape $x/y$ of the cheapest box.

✅ Correct!

y' = -2y/x

from the constraint turns

\dfrac{dC}{dx} = 6x + 4y + 4xy'

into

6x - 4y = 0

, so

y = \tfrac{3}{2}x

: an expensive bottom makes the box narrower and taller.

❌ That ignores the price of the bottom. The bottom term is now

3x^2

, whose derivative is

6x

, not

2x

❌ That is $y/x$ .

6x = 4y

gives

y = \tfrac{3}{2}x

, so

x/y = \tfrac{2}{3}

❌ The cost factor does not multiply the old ratio. A costlier bottom should shrink the base, not widen it; redo

\dfrac{dC}{dx} = 0

with

y' = -2y/x

❌ Not quite. Differentiate

V = x^2 y

for

y'

, then differentiate

C

and substitute.

Show solution

Constraint: treat $y = y(x)$ and differentiate $V = x^2 y$ :

0 = 2xy + x^2 y' \;\Rightarrow\; y' = -\frac{2xy}{x^2} = -\frac{2y}{x}.

Cost: by the product rule on $4xy$ ,

\frac{dC}{dx} = 6x + 4y + 4x y' = 6x + 4y + 4x\left(-\frac{2y}{x}\right) = 6x + 4y - 8y = 6x - 4y.

Setting $\dfrac{dC}{dx} = 0$ : $6x = 4y$ , so $\dfrac{x}{y} = \dfrac{4}{6} = \dfrac{2}{3}$ .

As for the area, $C \to \infty$ at both ends of $0 < x < \infty$ (substitute $y = V/x^2$ : $C = 3x^2 + 4V/x$ ), so this is the minimum. With the sides and the bottom equally priced, the same steps give $2x - 4y = 0$ and the familiar $x/y = 2$ .

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