Question

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

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

Short answer

Maximum & minimum value are\(\frac{3}{2}{\rm{ and }}\frac{1}{2}\)respectively.

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\(f\).

Given data

Consider the function\(f(x,y,z) = yz + xy\)

The constraints are\(xy = 1,{y^2} + {z^2} = 1\)

The given data can be rewritten as:

\(f(y,z) = yz + 1\)subject to the single constraint\(g(y,z) = {y^2} + {z^2} = 1\).

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

Substituting expression for\(f\,\& \,g\)we have:

\(\begin{array}{l}\nabla (yz + 1) = \lambda \nabla ({y^2} + {z^2} - 1)\\z + y = \lambda (2y + 2z)\end{array}\)

Comparing both sides:

\(\begin{array}{l}z = 2\lambda y\;\;\; \ldots \ldots (1)\\y = 2\lambda z \ldots \ldots {\rm{ (2) }}\end{array}\)

Solve equation

Substitute equation (1) in (2) and solve for\(\lambda \).

\(\begin{array}{l}y = 2\lambda (2\lambda y)\\y = 4{\lambda ^2}y\\\lambda = \pm \frac{1}{2}\end{array}\)

Substitute the value of\(\lambda \)in equation (1):

\(z = \pm y\)

Find critical point

Substitute equation (4) into equation (3) and solve.

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

Therefore, the critical points are, \(\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right),\left( {\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right),\left( { - \frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right){\rm{, and }}\left( { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right){\rm{. }}\)

Find maximum & minimum value

Substitute value of\(z\,\& \,y\)in\(f(y,z) = yz + 1\), we get

\(\begin{array}{c}f\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right) = f\left( { - \frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right)\;\;\;\\ = \sqrt {\frac{1}{2}} \times \sqrt {\frac{1}{2}} + 1\\ = \frac{3}{2}\\f\left( {\sqrt {\frac{1}{2}} , - \sqrt {\frac{1}{2}} } \right) = f\left( { - \sqrt {\frac{1}{2}} ,\sqrt {\frac{1}{2}} } \right)\\ = \sqrt {\frac{1}{2}} \times \left\{ { - \sqrt {\frac{1}{2}} } \right\} + 1\\ = \frac{1}{2}\end{array}\)

Hence, the maximum and minimum values are\(\frac{3}{2}{\rm{ and }}\frac{1}{2}\)