Solve the problem.
-Find the maximum value of $f ( x , y , z ) = x + 2 y + 3 z$ subject to $x - y + z = 1$ and $x ^ { 2 } + y ^ { 2 } = 1$.
A) $3 + \sqrt { 29 }$
B) $3 - \sqrt { 37 }$
C) $3 - \sqrt { 29 }$
D) $3 + \sqrt { 37 }$

Question

Quizplus · Accepted Answer

We can use Lagrange multipliers to solve the problem. Let
$$\begin{aligned} \mathcal{L} & = x + 2 y + 3 z - \lambda_1 (x - y + z - 1) - \lambda_2(x^2 + y^2 - 1) \ & = (\lambda_1 - \lambda_2)x + (\lambda_2 - \lambda_1) y + (\lambda_1 + \lambda_2)z - \lambda_1 + \lambda_2 \end{aligned}$$
Setting the partial derivatives equal to zero, we have
$$\begin{aligned} \lambda_1 - \lambda_2 & = 1 \ \lambda_2 - \lambda_1 & = 2\lambda_1 x \ \lambda_1 + \lambda_2 & = 3\lambda_2 y \ {\lambda_1} &= 3\lambda_2 z \end{aligned}$$
Solving these equations, we get
$$\begin{aligned} x & = \frac{3}{\sqrt{29}} \ y & = \frac{1}{\sqrt{29}} \ z & = \frac{2}{\sqrt{29}} \ \lambda_1 & = \frac{5}{\sqrt{29}} \ \lambda_2 & = \frac{2}{\sqrt{29}} \end{aligned}$$
Substituting these values into the objective function, we get
$$f(x,y,z) = x + 2y + 3z = (3 + \sqrt{29})$$
Therefore, the maximum value of $f(x,y,z)$ is $3 + \sqrt{29}$, which is option $\boxed{	extbf{(A)}}$.

Solve the Problem f(x,y,z)=x+2y+3zf ( x , y , z ) = x + 2 y + 3 zf(x,y,z)=x+2y+3z

Solve the Problem $f ( x , y , z ) = x + 2 y + 3 z$