Question

Find the maximum and minimum values of f(x, y, z) = xyz subject to the constraint 4x²+9y+z² = 4 Assume that y ≥ 0. Why is this assumption required?

Answer 1

$\displaystyle \text{First we define a new a function F(x,y)}\\ F(x,y,z,\lambda) = xyz + \lambda(4x^2+9y+z^2)\\ \text{Next, we differentiate F with respect to x,y and z and equate them}\\ \text{to 0}\\ F_x(x,y,z,\lambda) = yz + 8x\lambda=0 \qquad -(1)\\ F_y(x,y,z,\lambda) = xz + 9\lambda=0 \qquad-(2)\\ F_z(x,y,z,\lambda) = xy+ 2z\lambda=0 \qquad -(3)\\ \text{Using elimination method, we obtain }\\ z=2x, y = \frac{8}{9}x^2\\ \text{Substituting in the given constraint, we have that}\\ x=\pm \frac{1}{2}, y= \pm \frac{2}{9}, \pm 1\\ f(\frac{1}{2},\frac{2}{9},1)=\frac{1}{9}, f(-\frac{1}{2},-\frac{2}{9},-1)=-\frac{1}{9}\\\text{Therefore, $\frac{1}{9}$ is an absolute maximum at point ($\frac{1}{2},\frac{2}{9},1)$ and}\\ \text{point $-\frac{1}{9}$ is an absolute maximum at point ($-\frac{1}{2},-\frac{2}{9},-1)$}$