Find the extreme values of the function subject to the given constraint.
-$$f ( x , y ) = 4 x + 6 y , x ^ { 2 } + y ^ { 2 } = 13$$
A) Maximum: 26 at (2, 3); minimum: 0 at (0, 0)
B) Maximum: 36 at (3, 4); minimum: 0 at (0, 0)
C) Maximum: 26 at (2, 3); minimum: -26 at (-2, -3)
D) Maximum: 36 at (3, 4); minimum: -36 at (-3, -4)

Question

Quizplus · Accepted Answer

We can use Lagrange Multipliers to find the extreme values of $f(x,y)$ subject to the constraint $x^2+y^2=13$. 
Let $g(x,y)=x^2+y^2$, then $\nabla f=\lambda \nabla g$ for some scalar $\lambda$. 
Thus, we have the following system of equations: 
\begin{align*}
4&=2\lambda x \
6&=2\lambda y \
x^2+y^2&=13 
\end{align*}
Solving for $x$ and $y$, we get $x=2\lambda$ and $y=3\lambda$. Substituting into the third equation, we get $13=13\lambda^2$. 
Thus, $\lambda=\pm 1$ and the corresponding values of $x$ and $y$ are $(2,3)$ and $(-2,-3)$. 
Plugging these values back into $f(x,y)$, we get 
$$f(2,3)=4(2)+6(3)=26$$ 
$$f(-2,-3)=4(-2)+6(-3)=-26$$ 
Thus, the maximum value of $f(x,y)$ is 26 at $(2,3)$ and the minimum value of $f(x,y)$ is -26 at $(-2,-3)$. Therefore, the answer is C.

Find the Extreme Values of the Function Subject to the Given