Question

This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. f(x, y, z, t) = x + y + z + t; x2 + y2 + z2 + t2 = 1

Answer 1

The Lagrangian is

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

with critical points wherever

$L_x=1+2\lambda x=0$

$L_y=1+2\lambda y=0$

$L_z=1+2\lambda z=0$

$L_t=1+2\lambda t=0$

$L_\lambda=x^2+y^2+z^2+t^2-1=0$

The first four equations tell us

$x=y=z=t=-\dfrac1{2\lambda}$

and substituting these into the last equation gives

$4\left(-\dfrac1{2\lambda}\right)^2=\dfrac1{\lambda^2}=1\implies\lambda=\pm1$

If $\lambda=1$, then $x=y=z=t=-\dfrac12$; if $\lambda=-\dfrac12$, then $x=y=z=t=\dfrac12$. We then find a minimum value of

$f\left(-\dfrac12,-\dfrac12,-\dfrac12,-\dfrac12\right)=-2$

and a maximum value of

$f\left(\dfrac12,\dfrac12,\dfrac12,\dfrac12\right)=2$