I have this problem:
On space $ (Oxyz)$ given point $M(1,2,3)$.
Plane ($\alpha$) contain point $M$ and ($\alpha$) cross $Ox$ at $A(a,0,0)$; $Oy$ at $B(0,b,0)$; $C(0,0,c)$.
Where a,b,c>0
Write the equation of plane ($\alpha$) such that It makes $V_{OABC}$ reach minimum.
I don't know which inequality should use in here to find out $\min_{V_{OAB}}$ .
Please help me.
P\s: We have formula: $V_{OABC} = \frac{1}{6}a.b.c$
On
Suppose $A$, $B$, $C$ are in their volume-minimizing locations. Let $M_x$ be the point where $\overleftrightarrow{AM}$ meets $\overline{BC}$. (Define $M_y$ and $M_z$ similarly.) Since $V = \frac{1}{3} a \cdot |\triangle BOC|$, and since we've assumed $V$ is minimized, it must be that $\overline{BC}$ is the segment through $M_x$ that minimizes the area of $\triangle BOC$. Consequently, $\overline{BC}$ is the particular segment that has $M_x$ as its midpoint. (Why? (No calculus necessary!)) Likewise, $\overline{CA}$ has midpoint $M_y$, and $\overline{AB}$ has midpoint $M_z$. This says that $\overline{AM_x}$, $\overline{BM_y}$, $\overline{CM_z}$ are medians of $\triangle ABC$, so that their common point, $M$, is the triangle's centroid. Hence, $(1,2,3) = M = \frac{1}{3}\left(A+B+C\right) = \left(\frac{1}{3}a, \frac{1}{3}b, \frac{1}{3}c\right)$, and we conclude that $a=3$, $b=6$, $c=9$.
On
The plane through $(a,0,0)$, $(0,b,0)$, and $(0,0,c)$ has equation $$ \frac1a x + \frac1b y + \frac1c z = 1 $$ So the problem is:
Minimize $\frac16 abc$ subject to the constraint $\frac1a+\frac2b+\frac3c=1$.
By the AM/GM inequality, $$ \sqrt[3]{\frac{1\cdot2\cdot3}{a\cdot b\cdot c}} \le \frac13\left(\frac1a+\frac2b+\frac3c\right) = \frac13 $$ Rearranging yields $\frac16 abc\ge 27$, and we have equality when $\frac1a=\frac2b=\frac3c$, which yields $a=3$, $b=6$, $c=9$, as in the other answers.
Your plane has the equation
$$bc\,x + ac\,y + ab\,z = abc$$
Plugging $M$ into that, you get
$$bc + 2ac + 3ab = abc$$
With this, you can eliminate one variable, e.g. c
$$ (ab - b - 2a)c = 3ab \qquad c=\frac{3ab}{ab-b-2a}$$
Here you can see that $c>0$ translates to
$$ ab > b + 2a $$
That expression for $c$ plugged into the volume formula gives you
$$ V = \frac{abc}{6} = \frac{a^2b^2}{2(ab-b-2a)} $$
Now compute partial derivatives
\begin{align*} \frac{\partial V}{\partial a} &= -\frac{a^{2} {\left(b - 2\right)} b^{2}}{2 \, {\left({\left(a - 1\right)} b - 2 \, a\right)}^{2}} + \frac{a b^{2}}{{\left(a - 1\right)} b - 2 \, a} \\ \frac{\partial V}{\partial b} &= -\frac{{\left(a - 1\right)} a^{2} b^{2}}{2 \, {\left({\left(a - 1\right)} b - 2 \, a\right)}^{2}} + \frac{a^{2} b}{{\left(a - 1\right)} b - 2 \, a} \end{align*}
Setting both of these equal zero gives a critical point. Solving the resulting non-linear system of equations gives two solutions which don't agree with $a,b,c>0$, namely $a=0$ and $b=0$. The third solution will therefore be the minimum:
$$ a=3 \qquad b=6 \qquad c=9 $$
That it is a minimum and not a maximum can be seen from the fact that if the plane becomes more and more parallel to one of the coordinate axes, the volume will become arbitrarily big, so there can be no (global) maximum. You could also look at the second derivatives:
\begin{align*} \frac{\partial^2 V}{\partial a^2} &= \frac{a^{2} {\left(b - 2\right)}^{2} b^{2}}{{\left(a b - 2 \, a - b\right)}^{3}} - \frac{2 \, a {\left(b - 2\right)} b^{2}}{{\left(a b - 2 \, a - b\right)}^{2}} + \frac{b^{2}}{a b - 2 \, a - b} = 6 \\ \frac{\partial^2 V}{\partial b^2} &= \frac{{\left(a - 1\right)}^{2} a^{2} b^{2}}{{\left(a b - 2 \, a - b\right)}^{3}} - \frac{2 \, {\left(a - 1\right)} a^{2} b}{{\left(a b - 2 \, a - b\right)}^{2}} + \frac{a^{2}}{a b - 2 \, a - b} = \frac32 \end{align*} The corresponding equation for the plane is
\begin{align*} 54x + 27y + 18z &= 162 \\ 6x + 3y + 2z &= 18 \end{align*}