# I have three vectors barx=a hati+b hatj+c hat{k}, bary=b hati+c hatj+a hatk, barz=chati+ahatj+bhatk

Finding the maximum possible volume of a tetrahedron
Suppose I have three vectors $\overline{x}=a\stackrel{^}{i}+b\stackrel{^}{j}+c\stackrel{^}{k}$, $\overline{y}=b\stackrel{^}{i}+c\stackrel{^}{j}+a\stackrel{^}{k}$, $\overline{z}=c\stackrel{^}{i}+a\stackrel{^}{j}+b\stackrel{^}{k}$, where we can choose a, b, c from the set {1,2,3,....13}. We have to find the probability of the tetrahedron formed by $\overline{x}$, $\overline{y}$, $\overline{z}$ having maximum volume. How should I maximise the determinant for finding the maximum possible volume of the resulting tetrahedron? Can someone help me figure this out?
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Sheldon Castillo
Step 1
Start by expanding the function for the volume of the tetrahedron(up to a factor of $±1$, disregarding the absolute value for now to make calculations easier) which we know is 1/6:th of the corresponding parallelepiped:
$\mathbit{V}\left(a,b,c\right)=1/6|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}|=-\frac{1}{6}\left({a}^{3}+{b}^{3}+{c}^{3}-3abc\right)$
Step 2
Now even though this function only inputs and outputs discrete values, we may take partial derivatives, and set them equal to zero. However solving that system of equations gives two solutions which both of trivially give the volume of zero, so they are indeed minima. Notice that these were found without any constraints. We are looking for the maximum value for the function in the set $\left\{\left(a,b,c\right)|a,b,c\in \left\{1,2,...,13\right\}\right\}$. We can conclude that the maximum values for the volume are found on the vertices of this "cube". Plugging in $|\mathbit{V}\left(1,13,13\right)|=|\mathbit{V}\left(13,1,13\right)|=|\mathbit{V}\left(13,13,1\right)|=|\mathbit{V}\left(1,13,1\right)|=648$ rest of the vertices give a volume of either 0 or 360.
So in the end a total of 4 cases of reaching the maximum value, so the probability that the tetrahedron gets its maximum value is $\frac{4}{{13}^{3}}$