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Find an equation of the plane that passes through the point $(1,2,3)$, and cuts off the smallest volume in the first octant.

This is what i've done so far....

Let $a,b,c$ be some points that the plane cuts the $x,y,z$ axes. --> $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where $a,b,c >0$.

I saw a solution for this question was to use Lagrange multiplier. The solution goes as follows...

The product $abc$ will be equal to $6$ times the volume of the tetrahedron $OABC$ (could someone explain to my why is this so?)

$f(a,b,c) = abc$ given the condition $(\frac1a + \frac2b + \frac3b -1)$

$f(a,b,c) = abc + \lambda (\frac1a + \frac2b + \frac3c -1)$

2nd query to the question...

$f_a = \lambda g_a \Rightarrow bc - \frac\lambda {a^2} ; a = \sqrt \frac \lambda {bc} \\f_b = \lambda g_b \Rightarrow ac - \frac\lambda {b^2} ; b = \sqrt \frac {2\lambda}{ac} \\f_c = \lambda g_c \Rightarrow ab - \frac\lambda {c^2} ; c = \sqrt \frac {3\lambda}{ab}$

using values of $a,b,c$ into $\frac1a+\frac1b+\frac1c = 1\Rightarrow \lambda =\frac{abc}{a+2b+3c}$.

May i know how should i proceed to solve the unknowns?

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Do i let b,c = 0 to find a? any help?? – melyong Nov 12 '12 at 11:19
up vote 2 down vote accepted

The volume of a pyramid (of any shaped base) is $\frac13A_bh$, where $A_b$ is the area of the base and $h$ is the height (perpendicular distance from the base to the opposing vertex). In this particular case, we're considering a triangular pyramid, with the right triangle $OAB$ as a base and opposing vertex $C$. The area of the base is $\frac12ab$, and the height is $c$, so the volume of the tetrahedron is $\frac16abc$--equivalently, $abc$ is $6$ times the volume of the tetrahedron.

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Thank you Cameron! You clearly explained my doubt :D I have a clearer picture in mind now :D – melyong Nov 11 '12 at 18:44

The area of the right triangle with legs $a$ and $b$ is $(1/2)ab$. By integration, the volume of a generalized cone with base area $A$ and height $c$ is $(1/3)Ac$.

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Imagine a cube of side length $s$. It has 6 faces. Use each of these faces as the base of a pyramid whose apex is at the epicenter of the cube.

So now you have a square pyramid (six actually) whose volume is $\frac{1}{6}s\cdot s\cdot s$. Since the height of this thing is $\frac{1}{2}s$, you have a square pyramid whose volume is $\frac{1}{3}h\cdot s\cdot s$.

Now if you sliced it vertically in half using a plane that cuts through the apex and two opposite sides of the square, you have a pyramid with a right triangular base whose volume is $\frac{1}{6}h\cdot s\cdot s$.

It's not a right pyramid though, because the apex is not directly above the right angle in the triangle. No matter: horizontal slices of the pyramid can be shifted horizontally, and not affect the volume. So we can shift horizontal slices to form a true right pyramid (simplex) and still have volume $\frac{1}{6}h\cdot s\cdot s$

Lastly, any of the three orthogonal dimensions can be rescaled (stretched or compressed), and the net effect on volume will be to scale it by the same amount. (If volume is just a large sum of volumes of very tiny rectangular cells, then this should be understandable.) So we could scale in the various dimensions and have a right triangular pyramid whose volume is $\frac{1}{6}h\cdot b\cdot w$.

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thank you alex for your detailed explanation :) – melyong Nov 11 '12 at 18:55

BTW: you don't need Lagrange multipliers for this problem:

To find the minimum of $abc$ for $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}=1$ you can use $AM-GM$:

$$\frac{1}{3}=\frac{\frac{1}{a}+\frac{2}{b}+\frac{3}{c}}{3} \geq \sqrt[3]{\frac{6}{abc}}$$


$$abc \geq 3^3 \cdot 6 \,,$$ with equality if and only if


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Could I ask how to did u get $acb \Rightarrow \frac 1a + \frac 2b + \frac 3c = 1$ ?? – melyong Nov 15 '12 at 5:35
The equation of your plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, and the point $(1,2,3)$ is in the plane... – N. S. Nov 15 '12 at 5:38

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