Multivariable-Calculus · Unit 12 · Video 3 · Interactive Practice

The Least-Area Gold Pyramid: Its Top Sits Over the Incenter

IKey Formulas

Formula	Name	Where it comes from
$h = \dfrac{3V}{A_{\text{base}}}$	Height is fixed	$V = \tfrac{1}{3}A_{\text{base}}\,h$ with $V$ and $A_{\text{base}}$ given, so the apex moves in the plane $z = h$
$f(u_1, u_2, u_3) = \sum_{i=1}^{3} \tfrac{1}{2}a_i\sqrt{u_i^2 + h^2}$	Side area to minimize	Face $i$ has base $a_i$ and height $\sqrt{u_i^2 + h^2}$ , where $u_i$ is the distance from $Q$ to side $i$
$g(u_1, u_2, u_3) = \sum_{i=1}^{3} \tfrac{1}{2}a_i u_i = A_{\text{base}}$	Linear constraint	Joining $Q$ to the vertices cuts the base into three triangles of areas $\tfrac{1}{2}a_i u_i$
$\lambda = \dfrac{u_i}{\sqrt{u_i^2 + h^2}} \;\Rightarrow\; u_1 = u_2 = u_3 = r = \dfrac{2A_{\text{base}}}{a_1 + a_2 + a_3}$	Lagrange solution	$\nabla f = \lambda\nabla g$ after cancelling $\tfrac{a_i}{2}$ ; $\varphi(u) = u/\sqrt{u^2+h^2}$ is increasing, so equal values force equal distances

Key Insight: The apex goes directly above the incenter of the base. Every face then has height $\sqrt{r^2 + h^2}$ , so the least side area is $\tfrac{1}{2}(a_1 + a_2 + a_3)\sqrt{r^2 + h^2}$ .

IIWhere Do the Three Multipliers Agree?

Every $Q$ in the base satisfies $g = A_{\text{base}}$ ; only one makes the three ratios $u_i/\sqrt{u_i^2+h^2}$ equal.

Height h = 3V/A_base 3

💡 Each term of $f$ is a convex function of its own $u_i$ and $g$ is linear, so the single solution of the Lagrange equations is the minimum, not a maximum.

IIIIncenter or Centroid?

The least side area $\tfrac{1}{2}(a_1+a_2+a_3)\sqrt{r^2+h^2}$ needs the incenter; above the centroid the gold costs more.

Base Height h 3

IVQuiz Questions

Problem 1 · The Least Side Area

Given: a pyramid stands on a right triangle with sides $3$ , $4$ , $5$ and has volume $V = 8$ — find the least possible total area of its three side faces.

✅ Correct!

h = 4

r = 1

, and each face has height

\sqrt{1 + 16} = \sqrt{17}

, so the side area is

\tfrac{1}{2}(12)\sqrt{17} = 6\sqrt{17} \approx 24.74

❌ Missing the one half. A face is a triangle: its area is

\tfrac{1}{2}a_i\sqrt{r^2+h^2}

, not

a_i\sqrt{r^2+h^2}

❌ Check the height.

V = \tfrac{1}{3}A_{\text{base}}h

, so

h = \tfrac{3V}{A_{\text{base}}} = \tfrac{24}{6} = 4

, not

\tfrac{V}{A_{\text{base}}} = \tfrac{4}{3}

❌ The height is not the volume. Solve

V = \tfrac{1}{3}A_{\text{base}}h

for

h

h = 4

❌ Not quite. Find

h

from the volume,

r = 2A_{\text{base}}/(a_1+a_2+a_3)

, then

\tfrac{1}{2}(a_1+a_2+a_3)\sqrt{r^2+h^2}

Show solution

The base area is $A_{\text{base}} = \tfrac{1}{2}(3)(4) = 6$ and the perimeter is $3 + 4 + 5 = 12$ .

h = \frac{3V}{A_{\text{base}}} = \frac{3(8)}{6} = 4, \qquad r = \frac{2A_{\text{base}}}{a_1 + a_2 + a_3} = \frac{12}{12} = 1

At the minimum the apex is above the incenter, so $u_1 = u_2 = u_3 = r = 1$ and every face has height $\sqrt{1^2 + 4^2} = \sqrt{17}$ :

f_{\min} = \tfrac{1}{2}(3 + 4 + 5)\sqrt{17} = 6\sqrt{17} \approx 24.74

Problem 2 · Which Center?

Given: the base triangle has vertices $(0, 0)$ , $(4, 0)$ , $(0, 3)$ in the $xy$ -plane and the volume is fixed — find the point $Q$ directly below the apex of the least-side-area pyramid.

✅ Correct! The incenter of this right triangle is

(r, r)

with

r = \tfrac{2(6)}{12} = 1

: it is distance

1

from all three sides.

❌ That is the centroid. The centroid is where the medians meet; its distances to the sides are

1

\tfrac{4}{3}

and

\tfrac{4}{5}

, which are unequal. The Lagrange equations force equal distances.

❌ That is the circumcenter (the midpoint of the hypotenuse), which lies on a side: its distance to the hypotenuse is

0

, not equal to the other two.

❌ Check the radius. The constraint gives

\tfrac{1}{2}u(a_1+a_2+a_3) = A_{\text{base}}

, so

u = \tfrac{2A_{\text{base}}}{12} = 1

, not

\tfrac{A_{\text{base}}}{12} = \tfrac{1}{2}

❌ Not quite. The minimizer is equidistant from the three sides.

Show solution

The Lagrange equations give $u_1 = u_2 = u_3$ , so $Q$ is the incenter. Here the sides have lengths $4$ , $3$ , $5$ , the area is $6$ , and

r = \frac{2A_{\text{base}}}{a_1 + a_2 + a_3} = \frac{12}{12} = 1.

A point at distance $1$ from both legs (the $x$ -axis and the $y$ -axis) is $(1, 1)$ . Check the hypotenuse $3x + 4y = 12$ : its distance from $(1, 1)$ is $\dfrac{|3 + 4 - 12|}{5} = 1$ ✓.

The centroid $\left(\tfrac{4}{3}, 1\right)$ is a different point: its distance to the $y$ -axis is $\tfrac{4}{3} \ne 1$ .

Problem 3 · The Multiplier at the Optimum

Given: the base is a $5$ - $12$ - $13$ right triangle and the volume is $V = 60$ — find the common distance $r$ from $Q$ to the sides and the value of the multiplier $\lambda$ at the minimum.

What is $r$ ?

What is $\lambda$ ?

✅ Correct!

r = 2

h = 6

, and

\lambda = \tfrac{2}{\sqrt{4 + 36}} = \tfrac{1}{\sqrt{10}}

❌ Check $r$ .

A_{\text{base}} = \tfrac{1}{2}(5)(12) = 30

and the perimeter is

30

, so

r = \tfrac{2(30)}{30}

❌ Check $\lambda$ . After cancelling

\tfrac{a_i}{2}

\lambda = \tfrac{u_i}{\sqrt{u_i^2 + h^2}}

with

u_i = r

and

h = \tfrac{3V}{A_{\text{base}}}

. It is a ratio below

1

, with

r

(not

h

) on top.

Show solution

$A_{\text{base}} = \tfrac{1}{2}(5)(12) = 30$ and $a_1 + a_2 + a_3 = 5 + 12 + 13 = 30$ , so

r = \frac{2A_{\text{base}}}{a_1 + a_2 + a_3} = \frac{60}{30} = 2, \qquad h = \frac{3V}{A_{\text{base}}} = \frac{180}{30} = 6.

The equation $\dfrac{\partial f}{\partial u_i} = \lambda\dfrac{\partial g}{\partial u_i}$ reads $\dfrac{a_i}{2}\dfrac{u_i}{\sqrt{u_i^2 + h^2}} = \lambda\dfrac{a_i}{2}$ , so

\lambda = \frac{r}{\sqrt{r^2 + h^2}} = \frac{2}{\sqrt{4 + 36}} = \frac{2}{2\sqrt{10}} = \frac{1}{\sqrt{10}} \approx 0.316.

Problem 4 · A Square Base

Given: a pyramid on a square base of side $2$ with volume $V = 4$ . The same variables work: $u_1, \dots, u_4$ are the distances from $Q$ to the four sides — find the least total area of the four side faces.

✅ Correct! All four distances equal

1

, the apex is above the center,

h = 3

, and the side area is

\tfrac{1}{2}(8)\sqrt{1 + 9} = 4\sqrt{10}

❌ Missing the one half. Each face is a triangle with area

\tfrac{1}{2}a_i\sqrt{u_i^2 + h^2}

❌ Wrong distance.

u_i

is the perpendicular distance from

Q

to a side (

1

), not to a corner (

\sqrt{2}

❌ Check the height.

h = \tfrac{3V}{A_{\text{base}}} = \tfrac{12}{4} = 3

, not

\tfrac{V}{A_{\text{base}}} = 1

❌ Not quite. Repeat the triangle argument with four faces.

Show solution

The height is $h = \tfrac{3V}{A_{\text{base}}} = \tfrac{3(4)}{4} = 3$ . With four sides of length $2$ :

f = \sum_{i=1}^{4} \tfrac{1}{2}(2)\sqrt{u_i^2 + 9}, \qquad g = \sum_{i=1}^{4} \tfrac{1}{2}(2)u_i = 4.

Cancelling $\tfrac{a_i}{2}$ again gives $\lambda = \dfrac{u_i}{\sqrt{u_i^2 + 9}}$ for $i = 1, \dots, 4$ , and since $\varphi(u) = u/\sqrt{u^2 + 9}$ is increasing, $u_1 = u_2 = u_3 = u_4 = u$ . The constraint gives $4u = 4$ , so $u = 1$ : $Q$ is the center of the square.

f_{\min} = \tfrac{1}{2}(8)\sqrt{1^2 + 3^2} = 4\sqrt{10} \approx 12.65

A square has a point equidistant from all four sides; a general quadrilateral does not, so there the equal-distance solution is unavailable and the minimum lies elsewhere.

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