Question

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

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

Short answer

Maximum: \(f(2,1,0) = 5\)

Minimum: \(f(0 - 1,0) = 1\)

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\(\nabla f(x,y,z) = \lambda \nabla g(x,y,z)g(x,y,z) = k\)

And

(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\(f\).

Use Lagrange multiplier

Consider the function\(f(x,y,z) = {x^2} + {y^2} + {z^2}\).

Let \(g(x,y,z) = x - y\)and\(h(x,y,z) = {y^2} - {z^2}\).

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

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

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

Comparing both sides:

\(\begin{array}{c}2x = \lambda \ldots \ldots (1)\\2y = - \lambda + 2\mu y......(2)\\2z = - 2\mu z \ldots \ldots (3)\\x - y = 1 \ldots \ldots (4)\\{y^2} - {z^2} = 1 \ldots \ldots (5)\end{array}\)

Solve equation

Solve equation (3).

\(\begin{array}{l}2z = - 2\mu z\\2z(1 + \mu ) = 0\\z = 0\,or\mu = - 1\end{array}\)

Put \(\mu = - 1\) in equation (2) we have:

\(\begin{array}{l}2y = - \lambda - 2y\\4y = - \lambda \\y = - \frac{\lambda }{4}\end{array}\)

Find value of variable

Substitute \(x = \frac{\lambda }{2},y = - \frac{\lambda }{4}\) in \(x - y = 1\).

\(\begin{array}{c}\frac{\lambda }{2} + \frac{\lambda }{4} = 1\\\frac{{2\lambda + \lambda }}{4} = 1\\\lambda = \frac{4}{3}\end{array}\)

Solving we have: \(x = \frac{2}{3},y = - \frac{1}{3}\)

But these values are not acceptable because \(y = - \frac{1}{3}\)gives imaginary value for\(z\).

Find value of variable

For \(z = 0\)

Substituting\(z = 0\)in equation (5) we have:

\(y = \pm 1\)

And\(y = \pm 1\) on substituting in equation (4) gives

\(x = 2,0\)

So, the points are\((2,1,0)\,\,\& \,(0, - 1,0)\).

Evaluate function at points

Plugging the points\((2,1,0)\) and\((0, - 1,0)\)into\(f(x,y,z)\).

\(\begin{array}{l}f(2,1,0) = 5\\f(0, - 1,0) = 1\end{array}\).