Question

\({\bf{1}} - {\bf{12}}\) Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint.

\(\;f(x,y,z) = {x^4} + {y^4} + {z^4};\;\;\;{x^2} + {y^2} + {z^2} = 1\)

Short answer

Maximum value is\(1\)and minimum value is\(\frac{1}{3}\).

Step-by-step solution

Method of Lagrange multipliers

To find the maximum and minimum values of\(f(x,y,z)\)subject to the constraint\(g(x,y,z) = k\)(assuming that these extreme values exist and\(\nabla g \ne {\bf{0}}\)on the surface\(g(x,y,z) = k)\):

(a) Find all values of\(x,y,z\), and\(\lambda \)such that\(\nabla f(x,y,z) = \lambda \nabla g(x,y,z)g(x,y,z) = k\)

(b) Evaluate\(f\)at all the points\((x,y,z)\)that result from step (a). The largest of these values is the maximum value of\(f\); the smallest is the minimum value of\(f\).

Evaluate\(\nabla f = \lambda \nabla g\)

Given \(f(x,y,z) = {x^4} + {y^4} + {z^4};\;\;g(x,y,z):\;{x^2} + {y^2} + {z^2} - 1 = 0\)

\(\nabla f = \lambda \nabla g\)

\(\begin{array}{l} \Rightarrow \nabla \left( {{x^4} + {y^4} + {z^4}} \right) = \lambda \nabla \left( {{x^2} + {y^2} + {z^2} - 1} \right)\\ \Rightarrow 4{x^3} + 4{y^3} + 4{z^3} = \lambda \left( {2x + 2y + 2z} \right)\end{array}\)

Comparing both sides:

\(\begin{array}{l}4{x^3} = 2\lambda x\,\,\, - - - (1)\\4{y^3} = 2\lambda y\,\, - - - (2)\\4{z^3} = 2\lambda z\,\,\, - - - (3)\end{array}\)

Solve equation

By equation (1) we have

\(\begin{array}{l}4{x^3} = \lambda 2x\\2{x^3} = \lambda x\\x\left( {2{x^2} - \lambda } \right) = 0\\x = 0{\rm{ or }}\lambda = 2{x^2}\,\,\,\,\,.....(4)\end{array}\)

By equation (2) we have

\(y = 0{\rm{ or }}\lambda = 2{y^2}\,\,\,\,.....(5)\)

Similarly, solve the equation (3)

\(z = 0{\rm{ or }}\lambda = 2{z^2}\,\,\,.....(6)\)

Find critical point

By equation (4), (5) & (6) we have:

\(\begin{array}{l}2{x^2} = 2{y^2} = 2{z^2}\\{x^2} = {y^2} = {z^2}\end{array}\)

Substitute in equation\({x^2} + {y^2} + {z^2} - 1 = 0\)we have:

\(\begin{array}{c}{x^2} + {y^2} + {z^2} = 1\\3{x^2} = 1\\x = \pm \frac{1}{{\sqrt 3 }}\end{array}\)

The possible points are\(\left( { \pm \frac{1}{{\sqrt 3 }}, \pm \frac{1}{{\sqrt 3 }}, \pm \frac{1}{{\sqrt 3 }}} \right)\)

Find critical points

Consider\(x = 0,y = 0\)

\(\begin{array}{c}{(0)^2} + {(0)^2} + {(z)^2} = 1\\z = \pm 1\end{array}\)

The possible points are\((0,0, \pm 1)\).

Similarly, the possible points are\(( \pm 1,0,0)\)and\((0, \pm 1,0)\).

Find critical point

For\(x = 0\)and\(\lambda = 2{y^2} = 2{z^2}\)in\({x^2} + {y^2} + {z^2} - 1 = 0\)we have:

\(\lambda = 1\)

From (5) & (6), \(2{y^2} = 2{z^2} = 1\)

Possible points are\(\left( {0, \pm \frac{1}{{\sqrt 2 }}, \pm \frac{1}{{\sqrt 2 }}} \right)\).

Find critical point

Put\(y = 0,z = 0\), the possible points are\(\left( { \pm \frac{1}{{\sqrt 2 }},0, \pm \frac{1}{{\sqrt 2 }}} \right),\left( { \pm \frac{1}{{\sqrt 2 }}, \pm \frac{1}{{\sqrt 2 }},0} \right)\).

Therefore, all the possible points are:

\(\begin{array}{l}\left( { \pm \frac{1}{{\sqrt 3 }}, \pm \frac{1}{{\sqrt 3 }}, \pm \frac{1}{{\sqrt 3 }}} \right),(0,0, \pm 1),( \pm 1,0,0),(0, \pm 1,0),\\\left( {0, \pm \frac{1}{{\sqrt 2 }}, \pm \frac{1}{{\sqrt 2 }}} \right),\left( { \pm \frac{1}{{\sqrt 2 }},0, \pm \frac{1}{{\sqrt 2 }}} \right),\left( { \pm \frac{1}{{\sqrt 2 }}, \pm \frac{1}{{\sqrt 2 }},0} \right)\end{array}\).

Evaluate function at critical points

At \(\left( { \pm \frac{1}{{\sqrt 3 }}, \pm \frac{1}{{\sqrt 3 }}, \pm \frac{1}{{\sqrt 3 }}} \right)\)

\(\begin{array}{c}f(x,y,z) = {\left( { \pm \frac{1}{{\sqrt 3 }}} \right)^4} + {\left( { \pm \frac{1}{{\sqrt 3 }}} \right)^4} + {\left( { \pm \frac{1}{{\sqrt 3 }}} \right)^4}\\ = \frac{1}{9} + \frac{1}{9} + \frac{1}{9}\\ = \frac{1}{3}\end{array}\)

Evaluate function at critical points

The function value at the points\((0,0, \pm 1),( \pm 1,0,0)\)and\((0, \pm 1,0)\)is,

\(\begin{array}{l}f( \pm 1,0,0) = 1\\f(0, \pm 1,0) = 1\\f(0,0, \pm 1) = 1\end{array}\)

The function value at the points\(\left( {0, \pm \frac{1}{{\sqrt 2 }}, \pm \frac{1}{{\sqrt 2 }}} \right),\left( { \pm \frac{1}{{\sqrt 2 }},0, \pm \frac{1}{{\sqrt 2 }}} \right),\left( { \pm \frac{1}{{\sqrt 2 }}, \pm \frac{1}{{\sqrt 2 }},0} \right)\)is,

\(\begin{array}{c}f\left( {0, \pm \frac{1}{{\sqrt 2 }}, \pm \frac{1}{{\sqrt 2 }}} \right) = \frac{1}{4} + \frac{1}{4}\\ = \frac{1}{2}\end{array}\)

\(f\left( { \pm \frac{1}{{\sqrt 2 }},0, \pm \frac{1}{{\sqrt 2 }}} \right) = \frac{1}{2}\)

\(f\left( { \pm \frac{1}{{\sqrt 2 }}, \pm \frac{1}{{\sqrt 2 }},0} \right) = \frac{1}{2}\)

Therefore, the maximum value of the function is\(1\)and the minimum value is\(\frac{1}{3}\).