Question

\(3 - 14\)Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function.

\(f(x,y) = xy + \frac{1}{x} + \frac{1}{y}\)

Short answer

Minimum value at \(f(1,1) = 3\)

Step-by-step solution

Second derivative test

Suppose the second partial derivatives of\(f\)are continuous on a disk with center\({\bf{(a,b)}}\), and suppose that\({f_x}(a,b) = 0\)and\({f_y}(a,b) = 0\). Let

\(D = D(a,b) = {f_{xx}}(a,b){f_{yy}}(a,b) - {\left( {{f_{xy}}(a,b)} \right)^2}\)

(a) If\(D > 0\)and\({f_{xx}}(a,b) > 0\), then\(f(a,b)\)is a local minimum.

(b) If\(D > 0\)and\({f_{xx}}(a,b) < 0\), then\(f(a,b)\)is a local maximum.

(c) If\(D < 0\), then\(f(a,b)\)is not a local maximum or minimum.

Find partial derivative

The given function is\(f(x,y) = xy + \frac{1}{x} + \frac{1}{y}\).

\(\begin{aligned}{l}f(x,y) = xy + \frac{1}{x} + \frac{1}{y}\\{f_x}(x,y) = y - \frac{1}{{{x^2}}}\end{aligned}\)

\({f_y}(x,y) = x - \frac{1}{{{y^2}}}\)

Find critical points

Put \({f_x} = 0\)and\({f_y} = 0\)

\(\begin{aligned}{c}{f_x}(x,y) &= y - \frac{1}{{{x^2}}}\\y &= \frac{1}{{{x^2}}}\,\,\,\,\, - - - - (1)\end{aligned}\)

And

\(\begin{aligned}{c}{f_y}(x,y) &= x - \frac{1}{{{y^2}}}\\x &= \frac{1}{{{y^2}}}\,\,\,\,\, - - - (2)\end{aligned}\)

Solving equation (1) & (2) critical point is\(\left( {1,1} \right)\).

Compute second derivative

Now find\({f_{xx}},{f_{yy}}and\,{f_{xy}}\).

\(\begin{aligned}{c}f(x,y) &= xy + \frac{1}{x} + \frac{1}{y}\\{f_x}(x,y) &= y - \frac{1}{{{x^2}}}\\{f_{xx}}(x,y) &= \frac{2}{{{x^3}}}\\{f_{xy}}(x,y) &= 1\\{f_{yy}}(x,y) &= \frac{2}{{{y^3}}}\end{aligned}\)

Compute discriminant at critical point

The value of discriminant is:

\(\begin{aligned}{c}D(x,y) &= {f_{xx}}(x,y){f_{yy}}(x,y) - {\left( {{f_{xy}}(x,y)} \right)^2}\\ &= \left( {\frac{2}{{{x^3}}}} \right)\left( {\frac{2}{{{y^3}}}} \right) - {(1)^2}\end{aligned}\)

The value at critical point is:

\(\begin{aligned}{c}D(x,y) &= \left( {\frac{2}{{{x^3}}}} \right)\left( {\frac{2}{{{y^3}}}} \right) - {(1)^2}\\D(1,1) &= \left( {\frac{2}{{{1^3}}}} \right)\left( {\frac{2}{{{1^3}}}} \right) - {(1)^2}\\ &= 3 > 0\end{aligned}\)

Compute double derivative at critical point

Since\(D > 0\)the test also requires\({f_{xx}}(x,y)\),

So,

\(\begin{aligned}{c}{f_{xx}}(x,y) &= \frac{2}{{{x^3}}}\\{f_{xx}}(1,1) &= \frac{2}{{{x^3}}}\\ &= 2\end{aligned}\)

The function has minimum at,

\(\begin{aligned}{c}f(x,y) &= xy + \frac{1}{x} + \frac{1}{y}\\f(1,1) &= 1 + \frac{1}{1} + \frac{1}{1}\\ &= 3\end{aligned}\)

Hence, the function has minimum at\(f(1,1) = 3\).

Graph

Using graphing calculator the graph is: