Question

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

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

Short answer

Maximum value is\(1 + 2\sqrt 2 \)& minimum value is\(1 - 2\sqrt 2 \).

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\(f(x,y,z) = x + 2y\)

Constraints are given to be\(x + y + z = 1,{y^2} + {z^2} = 4\)

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

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:

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

Comparing both sides:

\(\begin{array}{l}1 = \lambda \ldots \ldots (1)\\2 = \lambda + 2\mu y \ldots \ldots (2)\\0 = \lambda + 2\mu z \ldots \ldots (3)\\1 = x + y + z \ldots \ldots (4)\\4 = {y^2} + {z^2} \ldots \ldots (5)\end{array}\)

Solve equation

Substitute\(1 = \lambda \)in equation (2), we get

\(\begin{array}{l}2 = 1 + 2\mu y\\1 = 2\mu y\\\mu = \frac{1}{{2y}} \ldots ..(6)\end{array}\)

Also substitute\(1 = \lambda \)in equation (3), we get

\(\begin{array}{l}0 = 1 + 2\mu z\\ - 1 = 2\mu z\\\mu = \frac{{ - 1}}{{2z}} \ldots ..(7)\end{array}\)

Solve equation

By equations\(\left( {{\rm{ }}6{\rm{ }}} \right){\rm{ }}and{\rm{ }}\left( {{\rm{ }}7} \right)\)we get

\(\begin{array}{l}\frac{1}{{2y}} = \frac{{ - 1}}{{2z}}\\y = - z\,\,\,....(8)\end{array}\)

Solve equations (8) and (4), we get

\(\begin{array}{l}1 = x + ( - z) + z\\x = 1\end{array}\)

Solving equations (8) and (5), we get

\(\begin{array}{l}{y^2} + {( - y)^2} = 4\\2{y^2} = 4\\y = \pm \sqrt 2 \end{array}\)

Find maximum & minimum value

Substitute\(y = \sqrt 2 \)in\(z = - y\) we have

\(z = - \sqrt 2 \)

For\(y = - \sqrt 2 \)then

\(z = \sqrt 2 \)

Substitute\(x = 1,y = \sqrt 2 \)and\(x = 1,y = - \sqrt 2 \)in\(f(x,y,z) = x + 2y\), we obtain\(f(1,\sqrt 2 , - \sqrt 2 ) = 1 + 2\sqrt 2 \)

And\(f(1, - \sqrt 2 ,\sqrt 2 ) = 1 - 2\sqrt 2 \)

Hence, the maximum and minimum values are\(1 + 2\sqrt 2 \)and\(1 - 2\sqrt 2 \).