Question

Use lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (if an answer does not exist, enter dne.) f(x, y, z, t) = x + y + z + t; x2 + y2 + z2 + t2 = 49

Answer 1

You have Lagrangian

$L(x,y,z,t,\lambda)=x+y+z+t+\lambda(x^2+y^2+z^2+t^2-49)$

with partial derivatives (set equal to 0)

$L_x=1+2\lambda x=0$
$L_y=1+2\lambda y=0$
$L_z=1+2\lambda z=0$
$L_t=1+2\lambda t=0$
$L_\lambda=x^2+y^2+z^2+t^2-49=0$

Solving the first four equations for $x,y,z,t$, respectively, we can substitute these solutions in terms of $\lambda$ into the fifth equation to find $\lambda$, which in turn will lead to $x,y,z,t$. Denoting by $\mathbf v=\begin{bmatrix}x&y&z&t\end{bmatrix}^\top$, we have

$1+2\lambda\mathbf v=0\implies\mathbf v=-\dfrac1{2\lambda}$

and substituting into the fifth equation yields

$\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=49$
$\dfrac1{\lambda^2}=49\implies\lambda=\pm\dfrac17$

From either choice of $\lambda$ we arrive at $x=y=z=t=\pm\dfrac72$, i.e. exactly two critical points at $\pm\left(\dfrac72,\dfrac72,\dfrac72,\dfrac72\right)$, for which we get a maximum value of 14 and minimum value of -14, respectively.