1. ## maximal volume question mmn 14 3B

V is the volume of a piramid which is created by the plane
$\frac{x}{\sqrt{x_0}}+\frac{y}{\sqrt{y_0}}+\frac{z} {\sqrt{z_0}}=\sqrt{a}$
and the x=0 y=0 z=0 planes

i cant imagine it
how its a piramid
i know that the volme is base multiplied by height/3

but here i just cant see how tho build the formula for it

in the solution they have done some sort of determinant divided by 6
dont know why

2. When y and z are both 0, that equation reduces to $x= \sqrt{ax_0}$. When x and z are both 0, that equation reduces to $y= \sqrt{ay_0}$. When x and y are both 0. that equation reduces to $z= \sqrt{az_0}$. So the plane contains the points $(\sqrt{ax_0}, 0, 0)$, $(0, \sqrt{ay_0}, 0)$ and $(0, 0, \sqrt{az_0})=$.

The "pyramid" has that as base and the xy, xz, and yz planes as sides with (0, 0, 0) as the vertex.

As far as the form of the answer in the book is concerned, they are probably using the fact that the volume of a parallelpiped, formed by three vectors, u, v, w, at one vertex is the "triple product", $u\cdot(v \times w)$. You probably know that the cross procuct of two vectors can be written as a "symbolic" determinant:
$\left|\begin{array}{ccc}i & j & k \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3\end{array}\right|$

The dot product of u with that is
$\left|\begin{array}{ccc}u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3\end{array}\right|$

It is not difficult to show, geometrically, that the volume of a "pyramid" (tetrahedron) formed by one corner of a paralellepiped is 1/6 of the volume of the parallelpiped.

3. its been solved by "lagrange conditions"
out condition is
$g(x)=\frac{x}{\sqrt{x_0}}+\frac{y}{\sqrt{y_0}}+\fr ac{z}{\sqrt{z_0}}-\sqrt{a}$

so the zero points are the points when the plane touches the axes.
i dont know what is vertex
but in wikipedia
vertex (plural vertices) is a special kind of point that describes the corners or intersections of geometric shapes.

the question states 4 planes cutting each other creating a paralellepiped some sort of cubic form.
and we need 3 vector comming out of the same vertex to find the volume of the paralellepiped

but our given plane in the formula can make only 2 vectors
because its 2d
i am not sure about the third one

its geometricly impossible to see how from one 2d plane we get 3 points of vectors on the 3d plane

4. its been solved by "lagrange conditions"
out condition is
$g(x)=\frac{x}{\sqrt{x_0}}+\frac{y}{\sqrt{y_0}}+\fr ac{z}{\sqrt{z_0}}-\sqrt{a}$

so the zero points are the points when the plane touches the axes.
i dont know what is vertex
but in wikipedia
vertex (plural vertices) is a special kind of point that describes the corners or intersections of geometric shapes.

the question states 4 planes cutting each other creating a paralellepiped some sort of cubic form.
and we need 3 vector comming out of the same vertex to find the volume of the paralellepiped

but our given plane in the formula can make only 2 vectors
because its 2d
i am not sure about the third one

its geometricly impossible to see how from one 2d plane we get 3 points of vectors on the 3d plane

maybe if the place is tilted downward than from 0,0,0 we get 3 vectors

but the 3d shape in this case is not a paralellepiped

??