Find the maximum and minimum values of the function \( f(x, y, z, t)=x+y+z+t \) subject to the constraint \( x^{2}+y^{2}+z^{2}+t^{2}=361 \). Maximum value is 1 Minimum value is occuring at 1

We wish to maximize and minimize \[ f(x,y,z,t)=x+y+z+t \] subject to the constraint \[ x^2+y^2+z^2+t^2=361. \] ### Step 1. Use the Lagrange Multipliers Method Introduce a Lagrange multiplier \(\lambda\) and consider the function \[ \mathcal{L}(x,y,z,t,\lambda)=x+y+z+t-\lambda \left(x^2+y^2+z^2+t^2-361\right). \] ### Step 2. Compute the Partial Derivatives Take the partial derivatives and set them equal to zero. For \(x\): \[ \frac{\partial \mathcal{L}}{\partial x}=1-2\lambda x=0 \quad \Longrightarrow \quad x=\frac{1}{2\lambda}. \] Similarly, for \(y\), \(z\), and \(t\) we obtain: \[ y=\frac{1}{2\lambda}, \quad z=\frac{1}{2\lambda}, \quad t=\frac{1}{2\lambda}. \] Thus, we have \[ x=y=z=t=c, \quad \text{where } c=\frac{1}{2\lambda}. \] ### Step 3. Apply the Constraint Substitute \(x=y=z=t=c\) into the constraint: \[ x^2+y^2+z^2+t^2=4c^2=361. \] Solve for \(c\): \[ 4c^2=361 \quad \Longrightarrow \quad c^2=\frac{361}{4} \quad \Longrightarrow \quad c= \pm \frac{19}{2}. \] ### Step 4. Find the Maximum and Minimum Values The function value is \[ f(x,y,z,t)=x+y+z+t=4c. \] - When \(c=\frac{19}{2}\): \[ f=4\left(\frac{19}{2}\right)=38. \] - When \(c=-\frac{19}{2}\): \[ f=4\left(-\frac{19}{2}\right)=-38. \] ### Final Answer The maximum value of \(f\) is \(38\), which occurs at \[ \left(\frac{19}{2},\frac{19}{2},\frac{19}{2},\frac{19}{2}\right). \] The minimum value of \(f\) is \(-38\), which occurs at \[ \left(-\frac{19}{2},-\frac{19}{2},-\frac{19}{2},-\frac{19}{2}\right). \]