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 $z=24$.