Question

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z = 1 and the cylinder x2 + y2 = 1. SOLUTION We maximize the function f(x, y, z) = x + 2y + 9z subject to the constraints g(x, y, z) = x − y + z = 1 and h(x, y, z) = x2 + y2 = 1. The Lagrange condition is ∇f = λ∇g + μ∇h, so we solve the equations

Answer 1

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$, so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$, we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$. The critical point with the negative $x$-coordinates gives the maximum value, $9+\sqrt{185}$.