Question

Find the extreme values of\(f\)subject to both constraints.

\(f(x,y,z) = 3x - y - 3z;\,\,x + y - z = 0,\;\;\;{x^2} + 2{z^2} = 1\)

Short answer

Maximum value is\(2\sqrt 6 \)and minimum value is\( - 2\sqrt 6 \).

Step-by-step solution

Lagrange multiplier condition

For two constraints\(\nabla f\left( {{x_0},{y_0},{z_0}} \right) = \lambda \nabla g\left( {{x_0},{y_0},{z_0}} \right) + \mu \nabla h\left( {{x_0},{y_0},{z_0}} \right)\).

Given data

Given function is\(f(x,y,z) = 3x - y - 3z{\rm{. }}\)

Constraints are\(g(x,y,z) = x + y - z = 0,\,\,h(x,y,z) = {x^2} + 2{z^2} = 1\)

By Lagrange's method of multipliers we find all \(x,y,z,\lambda \)and\(\mu \)such that

\(\vec \nabla f(x,y,z) = \lambda \vec \nabla g(x,y,z) + \mu \vec \nabla h(x,y,z)\)

Differentiate

Substituting values in\(\vec \nabla f(x,y,z) = \lambda \vec \nabla g(x,y,z) + \mu \vec \nabla h(x,y,z)\)we have:

\(\begin{array}{c}\vec \nabla (3x - y - 3z) = \lambda \vec \nabla (x + y - z) + \mu \vec \nabla h({x^2} + 2{z^2} - 1)\\3 - 1 - 3 = \lambda (1 + 1 - 1) + \mu (2x + 4z)\end{array}\)

Comparing both sides:

\(\begin{array}{l}3 = \lambda + 2\mu x\,\,\,\,\,....(1)\\ - 1 = \lambda \,\,\,...(2)\\ - 3 = - \lambda + 4\mu z\,\,\,....(3)\end{array}\)

Solve equation

Put\(\lambda = - 1\)into equation (1),

\(\begin{array}{l}3 = - 1 + \mu 2x\\4 = \mu 2x\\x = \frac{2}{\mu }\end{array}\)

Plugging\(\lambda = - 1\)into equation\((7)\) we have:

\(\begin{array}{l} - 3 = 1 + \mu 4z\\ - 4 = \mu 4z\\\frac{{ - 1}}{\mu } = z\end{array}\)

Substitution

Substitute values of\(x\,\& \,z\) in\({x^2} + 2{z^2} = 1\)we have

\(\begin{array}{c}{x^2} + 2{z^2} = 1\\{\left( {\frac{2}{\mu }} \right)^2} + 2{\left( {\frac{{ - 1}}{\mu }} \right)^2} = 1\\\frac{4}{{{\mu ^2}}} + \frac{2}{{{\mu ^2}}} = 1\\ \pm \sqrt 6 = \mu \end{array}\)

Find value of variables

The values of\(x\)and\(z\)at\(\mu = \sqrt 6 \)is,

\(x = \frac{2}{{\sqrt 6 }},z = - \frac{1}{{\sqrt 6 }}\)

The values of\(x\)and\(z\)at\(\mu = - \sqrt 6 \)is,

\(x = - \frac{2}{{\sqrt 6 }},z = \frac{1}{{\sqrt 6 }}\)

Find value of\({\bf{y}}\)

Substitute the values\(x = \frac{2}{{\sqrt 6 }}\)and\(z = - \frac{1}{{\sqrt 6 }}\)in\(x + y - z = 0\)

\(\begin{array}{c}\frac{2}{{\sqrt 6 }} + y + \frac{1}{{\sqrt 6 }} = 0\\y = - \frac{3}{{\sqrt 6 }}\end{array}\)

Substitute the values\(x = - \frac{2}{{\sqrt 6 }}\)and\(z = \frac{1}{{\sqrt 6 }}\)in\(x + y - z = 0\).

\(\begin{array}{*{20}{r}}{ - \frac{2}{{\sqrt 6 }} + y - \frac{1}{{\sqrt 6 }} = 0}\\{y = \frac{3}{{\sqrt 6 }}}\end{array}\)

Find maximum & minimum value

The value of\(f(x,y,z)\)at critical points is,

\(\begin{array}{c}f\left( {\frac{2}{{\sqrt 6 }},\frac{{ - 3}}{{\sqrt 6 }},\frac{{ - 1}}{{\sqrt 6 }}} \right) = 3\left( {\frac{2}{{\sqrt 6 }}} \right) - \left( {\frac{{ - 3}}{{\sqrt 6 }}} \right) - 3\left( {\frac{{ - 1}}{{\sqrt 6 }}} \right)\\ = 2\sqrt 6 \\f\left( {\frac{{ - 2}}{{\sqrt 6 }},\frac{3}{{\sqrt 6 }},\frac{1}{{\sqrt 6 }}} \right) = 3\left( {\frac{{ - 2}}{{\sqrt 6 }}} \right) - \left( {\frac{3}{{\sqrt 6 }}} \right) - 3\left( {\frac{1}{{\sqrt 6 }}} \right)\\ = - 2\sqrt 6 ({\rm{Minimum}})\end{array}\)

Hence, the minimum for the function is\( - 2\sqrt 6 \)and the maximum of the function is\(2\sqrt 6 \).