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I need to say which plane goes through the point (3,5,6), intersects with the positive side of all axis and creates the pyramid with the minimal volume with those axis.

I'm not sure how to begin this.

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Not quite sure I understand. Do you mean that the plane passes through (0,0,6), (0,5,0), and (3,0,0)? –  Neil Apr 26 '12 at 9:58
    
Here's an idea: can you figure out the general equation of a plane passing through that one point? You can determine where that plane crosses the axes easily from that. Then, you'll want to look into deriving the volume of an trirectangular tetrahedron... –  J. M. Apr 26 '12 at 9:59
    
To expand on that: A plane in $\mathbb{R}^3$ is uniquely defined by one point and a normal vector. Find constraints for your normal vector (intersections with axes). Then, within these constraints, find the normal vector that yields the minimal volume. –  m_l Apr 26 '12 at 10:05
    
still not sure I understand how to solve this.. can you give more concrete explanation please? –  Guy Apr 26 '12 at 10:21

2 Answers 2

up vote 3 down vote accepted

When the axis intercepts of your plane are $a$, $b$, $c>0$ then the volume of said pyramid is given by $V={1\over 6} abc$. The equation of this plane can be written in the form $${x\over a}+{y\over b}+{z\over c}=1\ ,$$ and as it has to pass through the point $(3,5,6)$ we have the extra condition $${3\over a}+{5\over b}+{6\over c}=1\ .\qquad(1)$$ Therefore we have to minimize $abc$ given the side condition $(1)$. This can be done with the help of Lagrange multipliers or with the help of a convexity argument.

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OP said, "I'm not sure how to begin this." I guess OP now knows how to begin, middle, and end this. –  Gerry Myerson Apr 26 '12 at 12:42

adding to what Christian Blatter said, we can observe that harmonic mean is less than geometric mean always, and the G.M == H.M only when all the numbers are equal. which would mean 3/a=1/3, 5/b=1/3 and 6/c=1/3

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