Question

Find the maximum and minimum values of\(f\)subject the given constraints. Use a computer algebra system to solve system of equations that arises in using Lagrange multiple. (If your CAS finds only one solution, you may need to additional commands.)

\(f(x,y,z) = y{e^{x - z}};\;\;\;9{x^2} + 4{y^2} + 36{z^2} = 36,xy + yz = 1\)

Short answer

Minimum value is \( - 5.3506\) & maximum value is\(9.7938\).

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\(f\).

Apply Lagrange’s multiplier

Consider \(f(x,y,z) = y{e^{x - z}},9{x^2} + 4{y^2} + 36{z^2} = 36,\;\;\;xy + yz = 1\)

Use Lagrange multipliers to optimize a function with two constraints.

\(\begin{array}{l}\nabla f = \lambda \nabla g + \mu \nabla h\\{f_x} = \lambda {g_x} + \mu {h_x}\\{f_y} = \lambda {g_y} + \mu {h_y}\\{f_z} = \lambda {g_z} + \mu {h_z}\\g(x,y,z) = k\\h(x,y,z) = c\end{array}\)

Form equation

By given data we have :

\(\begin{array}{l}f(x,y,z) = y{e^{x - z}}\\g(x,y,z) = 9{x^2} + 4{y^2} + 36{z^2}\\h(x,y,z) = xy + yz\end{array}\)

Substituting & comparing we get:

\(\begin{aligned}{c}y{e^x}{e^{ - z}} &= 18\lambda x + \mu y\\{e^x}{e^{ - z}} &= 8\lambda y + \mu (x + z)\\ - y{e^x}{e^{ - z}} &= 72\lambda z + \mu y\\36 &= 9{x^2} + 4{y^2} + 36{z^2}\\1 &= xy + yz\end{aligned}\)

Determine maximum & minimum

Use a computer algebra system to solve this system of equations to get four solutions:

\((x,y,z) = \left( {4z, \pm \sqrt {\frac{{45 \pm 3\sqrt {205} }}{{10}}} ,\frac{1}{{5y}}} \right)\)

Note that the domain is closed and bounded. Therefore\(f\)will obtain its maximum and minimum values. Substitute these four points into\(f\)to determine which is the maximum and which the minimum.

Using computer algebra system

\(\sqrt {\frac{{45 + 3\sqrt {205} }}{{10}}} {e^{\frac{4}{5}\sqrt {\frac{{10}}{{45 + 3\sqrt {205} }}} - \frac{1}{5}\sqrt {\frac{{10}}{{45 + 3\sqrt {205} }}} }} \approx 9.7938\)is maximum.

\(\sqrt {\frac{{45 - 3\sqrt {205} }}{{10}}} {e^{\frac{4}{5}\sqrt {\frac{{10}}{{45 - 3\sqrt {205} }}} }} - \frac{1}{5}\sqrt {\frac{{10}}{{45 - 3\sqrt {205} }}} \approx 1.70\)

\( - \sqrt {\frac{{45 + 3\sqrt {205} }}{{10}}} {e^{\frac{{ - 4}}{5}\sqrt {\frac{{10}}{{45 + 3\sqrt {205} }}} - \left( {\frac{{ - 1}}{5}\sqrt {\frac{{10}}{{45 + 3\sqrt {205} }}} } \right)}} \approx - 5.3506\)is minimum.

\( - \sqrt {\frac{{45 - 3\sqrt {205} }}{{10}}} {e^{\frac{{ - 4}}{5}\sqrt {\frac{{10}}{{45 - 3\sqrt {205} }}} \left( {\frac{{ - 1}}{5}\sqrt {\frac{{10}}{{45 - 3\sqrt {205} }}} } \right)}} \approx - 0.12\)