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 seek the minimum value of $z=12x+3y$ subject to the condition, or constraint, that $x$ and $y$ satisfy the equation $xy=4$. Since $x>0$ and $y>0,$ we can express this as a problem in one variable by solving $xy=4$ for $y$ and substituting $y=\dfrac{4}{x}$ in the expression $z=12x+3y.$ $$z=12x+3y \underset{\underset{\color{#0066A7}{\hbox{\(y=\dfrac{4}{x}\)}}}{\color{#0066A7}{\uparrow}}}{=}12x+\frac{12}{x} \qquad x>0$$

Then $$\frac{dz}{dx}=12-\frac{12}{x^{2}}=\dfrac{12x^{2}-12}{x^{2}}\qquad x>0$$

The critical numbers of $z$ are $-1$ and $1.$ We exclude $x=-1$, since $x>0$. To examine $x=1,$ we apply the Second Derivative Test. Since $\dfrac{d^{2}z}{dx^{2}}=\dfrac{24}{x^{3}}>0$ for $x=1$, we conclude that $z$ has a minimum when $x=1$ and $y=\dfrac{4}{x} =4$. The minimum value of $z$ is $24$ at the point $(1,4)$ on the hyperbola.