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${[yz-lambda_(1)-lambda_(2)=0],[xz-lambda_(1)+lambda_(2)=0],[xy-lambda_(2)=0],[-(x+y-2)=0]{[lambda_(2)=-lambda_(1)+yz=lambda_(1)-x_(2)],[lambda_(2)=lambda_(1)(-z+y)+y],[lambda_(2)=-xy]{lambda_(1):}$

$\left\{\begin{array}{l}y z-\lambda_{1}-\lambda_{2}=0 \\ x z-\lambda_{1}+\lambda_{2}=0 \\ x y-\lambda_{2}=0 \\ -(x+y-2)=0\end{array}\left\{\begin{array}{l}\lambda_{2}=-\lambda_{1}+y z=\lambda_{1}-x_{2} \\ \lambda_{2}=\lambda_{1}(-z+y)+y \\ \lambda_{2}=-x y\end{array}\left\{\lambda_{1}\right.\right.\right.$

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Math Problems

Calculus

Find derivatives of sine and cosine functions

Full solution

\left\{\begin{array}{l}y z-\lambda_{1}-\lambda_{2}=0 \\ x z-\lambda_{1}+\lambda_{2}=0 \\ x y-\lambda_{2}=0 \\ -(x+y-2)=0\end{array}\left\{\begin{array}{l}\lambda_{2}=-\lambda_{1}+y z=\lambda_{1}-x_{2} \\ \lambda_{2}=\lambda_{1}(-z+y)+y \\ \lambda_{2}=-x y\end{array}\left\{\lambda_{1}\right.\right.\right.

Solve for $\lambda_{2}$ : First, let's look at the third equation: $xy - \lambda_{2} = 0$ . We can solve for $\lambda_{2}$ : $\lambda_{2} = xy$ .
Substitute $\lambda_{2}$ : Now, let's use the first equation: $yz - \lambda_{1} - \lambda_{2} = 0$ . Substitute $\lambda_{2}$ with $xy$ : $yz - \lambda_{1} - xy = 0$ . Solve for $\lambda_{1}$ : $\lambda_{1} = yz - xy$ .
Solve for $\lambda_1$ : Next, we'll check the second equation: $xz - \lambda_1 + \lambda_2 = 0$ . Substitute $\lambda_1$ and $\lambda_2$ : $xz - (yz - xy) + xy = 0$ . Simplify: $xz - yz + xy + xy = 0$ . Oops, there's a mistake here. We should not have added $xy$ twice.