Find the point in the first quadrant on the hyperbola \(xy=4,\) where the value of \(z=12x+3y\) is a minimum. What is the minimum value?

Solution We can reword the problem to read as follows: Find the minimum value of \(z=f(x,y)=12x+3y\) subject to the condition \(xy=4,\) where \(x>0\) and \(y>0\).

The test points satisfy the equations \[ {\nabla }f(x,y)=\lambda {\nabla }g(x,y)\qquad g(x,y)=xy-4=0\qquad x>0\quad y>0 \]

where \(\lambda\) is a number.

893

Solve the system of equations \[ \left\{\begin{array}{l} 12 &=&\lambda y \qquad \enspace{\color{#0066A7}{\hbox{\(f_{x} (x,y)=\lambda g_{x} (x,y)\)}}}\\ 3 &=&\lambda x \qquad \enspace {\color{#0066A7}{\hbox{\(f_{y} (x,y)=\lambda g_{y} (x,y)\)}}}\\ xy-4 &=&0 \qquad \enspace \enspace {\color{#0066A7}{\hspace{.3pc}\hbox{\(g(x,y)=0\)}}} \end{array}\right. \]

Eliminating \(\lambda\) from the first two equations results in \(y=4x\), so the third equation becomes \begin{eqnarray*} &\qquad\qquad \enspace 4x^{2}-4=0 \\[2pt] &x=1 \qquad \enspace \enspace \hbox{or} \enspace \enspace \enspace x=-1 \end{eqnarray*}

Since \(x>0\), we ignore \(x=-1.\) When \(x=1\), then \(y=4\), so the only test point is \((1,4)\). The corresponding minimum value of \(z=12x+3y\) is \(z=24\).