Question

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

Short answer

Equation is\(6x + 3y + 2z = 18\).

Step-by-step solution

Formula

The equation of plane cutting the coordinate axes at\(a,b,c\)is given by\(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\).

Equation of plane

Let the required plane meets the coordinate axes at points\(A,B,C\) respective.

Given it passes through\((1,2,3)\).

Therefore, \(\frac{1}{a} + \frac{2}{b} + \frac{3}{c} = 1\)

\(\begin{array}{l}\frac{3}{c} = 1 - \frac{1}{a} - \frac{2}{b}\\\frac{3}{c} = \frac{{ab - b - 2a}}{{ab}}\\c = \frac{{3ab}}{{ab - b - 2a}}\end{array}\)

Find volume

Now, the plane forms tetrahedron in the first octant and the volume of tetrahedron

\(\begin{array}{c}V = \frac{1}{6}abc\\ = \frac{1}{6}ab\frac{{3ab}}{{ab - b - 2a}}\\ = \frac{{{a^2}{b^2}}}{{ab - b - 2a}}\end{array}\)

Differentiate

Differentiating \(\;v\)partially with respect to\(a\),

\(\begin{array}{c}\frac{{\partial v}}{{\partial a}} = \frac{\partial }{{\partial a}}\left( {\frac{{{a^2}{b^2}}}{{ab - b - 2a}}} \right)\\ = {b^2}\frac{\partial }{{\partial a}}\left( {\frac{{{a^2}}}{{(ab - b - 2a)}}} \right)\\ = \frac{{{b^2}\left( {(ab - b - 2a)\frac{\partial }{{\partial a}}{a^2} - {a^2}\frac{\partial }{{\partial a}}(ab - b - 2a)} \right)}}{{{{(ab - b - 2a)}^2}}}\\ = \frac{{a{b^2}(2ab - 2b - 4a - ab + 2a)}}{{{{(ab - b - 2a)}^2}}}\\ = \frac{{a{b^2}(ab - 2a - 2b)}}{{{{(ab - b - 2a)}^2}}}\end{array}\)

Differentiate

Differentiating \(v\)partially with respect to\(b\)

\(\begin{array}{c}\frac{{\partial v}}{{\partial b}} = \frac{\partial }{{\partial b}}\left( {\frac{{{a^2}{b^2}}}{{(ab - b - 2a)}}} \right)\\ = {a^2}\frac{\partial }{{\partial b}}\left( {\frac{{{b^2}}}{{ab - b - 2a}}} \right)\\ = \frac{{{a^2}\left( {(ab - b - 2a)\frac{\partial }{{\partial b}}{b^2} - \frac{\partial }{{\partial b}}(ab - b - 2a)} \right)}}{{{{(ab - b - 2a)}^2}}}\\ = \frac{{{a^2}\left( {(ab - b - 2a)2b - {b^2}(a - 1)} \right)}}{{{{(ab - b - 2a)}^2}}}\\\left. { = \frac{2}{{(ab - b - 2b}}} \right)\end{array}\)

Find critical points

For critical points, \(\frac{{\partial v}}{{\partial a}} = 0\)and\(\frac{{\partial v}}{{\partial b}} = 0\)

For\(\frac{{\partial v}}{{\partial a}} = 0\)gives,

\(\begin{array}{l}\frac{{a{b^2}(ab - 2a - 2b)}}{{{{(ab - b - 2a)}^2}}} = 0\\\;a = 0{\rm{ or }}b = 0{\rm{ or }}ab - 2a - 2b = 0\end{array}\)

Since \(a\)and\(b\)cannot be zero.

Therefore,

\(\begin{array}{c}ab - 2a - 2b = 0\\\;ab = 2a + 2b\\\frac{1}{a} + \frac{1}{b} = \frac{1}{2}\,\,\,\,\,\,\,\, - - - - (1)\end{array}\)

Find critical point

For\(\frac{{\partial v}}{{\partial b}} = 0\)we have:

\(\begin{array}{l}\frac{{{a^2}b(ab - 4a - b)}}{{ab - b - 2a}} = 0\\a = 0{\rm{ or }}b = 0{\rm{ or }}ab - 4a - b = 0\end{array}\)

Since \(a \ne 0,b \ne 0\)

Therefore,

\(\begin{array}{c}ab - 4a - b = 0\\ab = 4a + b\;\;\;\\\frac{1}{a} + \frac{4}{b} = 1\,\,\,\,\, - - - - (2)\end{array}\)

Using & solving equation (1) & (2) we have\(a = 3,b = 6,c = 9\).

Compute second order derivative

Now, differentiating \(\frac{{\partial v}}{{\partial a}}\)partially with respect to\(a\),

\(\begin{array}{c}\frac{{{\partial ^2}v}}{{\partial {a^2}}} = \frac{\partial }{{\partial a}}\frac{{a{b^2}(ab - 2a - 2b)}}{{{{(ab - b - 2a)}^2}}}\\ = {b^2}\frac{\partial }{{\partial a}}\frac{{\left( {{a^2}b - 2{a^2} - 2ab} \right)}}{{{{(ab - b - 2a)}^2}}}\\ = \frac{{{b^2}\left( {(ab - b - 2a)(2ab - 4a - 2b) - \left( {{a^2}b - 2{a^2} - 2ab} \right)2(b - 2)} \right)}}{{{{(ab - b - 2a)}^3}}}\end{array}\)

Therefore\(\frac{{{\partial ^2}v}}{{\partial {a^2}}}{\rm{ at }}a = 3{\rm{ and }}b = 6\)is\(12\).

Compute second order derivative

Differentiating \(\frac{{\partial v}}{{\partial a}}\)partially with respect to \(b\),

\(\begin{array}{c}\frac{{{\partial ^2}v}}{{\partial b\partial a}} = \frac{\partial }{{\partial b}}\left( {\frac{{\partial v}}{{\partial a}}} \right)\\a\frac{\partial }{{\partial b}}\frac{{\left( {a{b^3} - 2a{b^2} - 2{b^3}} \right)}}{{{{(ab - b - 2a)}^2}}}\\ = \frac{{a\left( {{{(ab - b - 2a)}^2}\frac{\partial }{{\partial b}}\left( {a{b^3} - 2a{b^2} - 2{b^3}} \right) - \left( {a{b^3} - 2a{b^2} - 2{b^3}} \right)\frac{\partial }{{\partial b}}{{(ab - b - 2a)}^2}} \right.}}{{{{(ab - b - 2a)}^4}}}\\ = \frac{{a\left( {(ab - b - 2a)\left( {3a{b^2} - 4ab - 6{b^2}} \right) - 2(a - 1)\left( {a{b^3} - 3a{b^2} - 2{b^3}} \right)} \right)}}{{{{(ab - b - 2a)}^3}}}\end{array}\)

Therefore at\({\rm{a}} = 3\)and \({\rm{b}} = 6,\frac{{{\partial ^2}v}}{{\partial b\partial a}} = 3\).

Compute second order derivative

Differentiating\(\frac{{\partial v}}{{\partial b}}\)partially with respect to \(b\),

\(\begin{array}{c}\frac{{{\partial ^2}v}}{{\partial {b^2}}} = \frac{\partial }{{\partial b}}\left( {\frac{{\partial v}}{{\partial b}}} \right)\\ = \frac{\partial }{{\partial b}}\frac{{{a^2}b(ab - 4a - b)}}{{{{(ab - b - 2a)}^2}}}\\ = \frac{{{a^2}\left( {(ab - b - 2a)\frac{\partial }{{\partial b}}\left( {a{b^2} - 4ab - {b^2}} \right) - \left( {a{b^2} - 4ab - {b^2}} \right)\frac{\partial }{{\partial b}}{{(ab - b - 2a)}^2}} \right)}}{{{{(ab - b - 2a)}^4}}}\\ = \frac{{{a^2}\left( {(2ab - 4a - 2b)(ab - b - 2a) - 2\left( {a{b^2} - 4ab - {b^2}} \right)(a - 1)} \right)}}{{{{(ab - b - 2a)}^3}}}\end{array}\)

At \({\rm{a}} = 3\)and \({\rm{b}} = 6,\frac{{{\partial ^2}v}}{{\partial {b^2}}} = 3\)

Compute discriminant

Now, \({\rm{D}}\)at\({\rm{a}} = 3\)and\({\rm{b}} = 6\)is:

\(\begin{array}{c}D = \left( {\frac{{{\partial ^2}v}}{{\partial {a^2}}}} \right)\left( {\frac{{{\partial ^2}v}}{{\partial {b^2}}}} \right) - {\left( {\frac{{{\partial ^2}v}}{{\partial b\partial a}}} \right)^2}\\ = 12 \times 3 - {(3)^2}\\ = 27 > 0\end{array}\)

Equation of plane

Volume\(v\) is minimum at\(a = 3,b = 6\)and\(c = 9\)equation of plane is,

\(\begin{array}{c}\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\\\;\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1\;\\\frac{{6x + 3y + 2z}}{{18}} = 1\;\\6x + 3y + 2z = 18\end{array}\)

Hence,

The equation of plane passing through\((1,2,3)\)nand cuts off the smallest volume in the first octant is\(6x + 3y + 2z = 18\)