Solve an Optimization Problem

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.