Find values of x, y, and $\lambda$ that satisfy the system.These systems arise in certain optimization problems in calculus, and $\lambda$ is called a Lagrange multiplier.
$\left\{\begin{array}{ll}
2 x+\lambda & =0 \
2 y+\lambda & =0 \
x+y-10 & =0
\end{array}\right.$
A)x = 5 , y = -10, $\lambda$ = -10
B) x = 5, y = 5, $\lambda$ = 5
C)x = -10, y = 5, $\lambda$ = -10
D)x = 5, y = -10, $\lambda$ = 5
E)x = 5, y = 5, $\lambda$ = -10

Question

Find values of x, y, and $\lambda$ that satisfy the system.These systems arise in certain optimization problems in calculus, and $\lambda$ is called a Lagrange multiplier.
 $\left\{\begin{array}{ll}
2 x+\lambda & =0 \
2 y+\lambda & =0 \
x+y-10 & =0
\end{array}\right.$  
A)x = 5 , y = -10, $\lambda$ = -10
B) x = 5, y = 5, $\lambda$ = 5
C)x = -10, y = 5, $\lambda$ = -10
D)x = 5, y = -10, $\lambda$ = 5
E)x = 5, y = 5, $\lambda$ = -10

Quizplus · Accepted Answer

From the first two equations, setting them equal to each other because they both equal $\lambda$, we get $2x = 2y$, which simplifies to $x = y$. Using the third equation $x + y = 10$, and substituting $y$ for $x$, we get $2x = 10$, so $x = 5$. Substituting $x = 5$ into $x + y = 10$, we get $y = 5$. Substituting $x = 5$ into the first equation $2x + \lambda = 0$, we get $10 + \lambda = 0$, so $\lambda = -10$.

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