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

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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

Answers

You have Lagrangian[tex]L(x,y,z,t,\lambda)=x+y+z+t+\lambda(x^2+y^2+z^2+t^2-49)[/tex]with partial derivatives (set equal to 0)[tex]L_x=1+2\lambda x=0[/tex][tex]L_y=1+2\lambda y=0[/tex][tex]L_z=1+2\lambda z=0[/tex][tex]L_t=1+2\lambda t=0[/tex][tex]L_\lambda=x^2+y^2+z^2+t^2-49=0[/tex]Solving the first four equations for [tex]x,y,z,t[/tex], respectively, we can substitute these solutions in terms of [tex]\lambda[/tex] into the fifth equation to find [tex]\lambda[/tex], which in turn will lead to [tex]x,y,z,t[/tex]. Denoting by [tex]\mathbf v=\begin{bmatrix}x&y&z&t\end{bmatrix}^\top[/tex], we have[tex]1+2\lambda\mathbf v=0\implies\mathbf v=-\dfrac1{2\lambda}[/tex]and substituting into the fifth equation yields[tex]\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=49[/tex][tex]\dfrac1{\lambda^2}=49\implies\lambda=\pm\dfrac17[/tex]From either choice of [tex]\lambda[/tex] we arrive at [tex]x=y=z=t=\pm\dfrac72[/tex], i.e. exactly two critical points at [tex]\pm\left(\dfrac72,\dfrac72,\dfrac72,\dfrac72\right)[/tex], for which we get a maximum value of 14 and minimum value of -14, respectively.

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