Answer to Question #107937 in Calculus for annie

Question

Find the points on the ellipse x^2 + 16y^2 = 16 that are furthest
away from the point (0, −1) using the method of Lagrange Multipliers.

Answer 1

when the distance from the point $(0,-1)$ is $d_{(x,y)}$ ,

problem is,

$d_{(x,y)}=(x-0)^2+(y-(-1))^2\\ d_{(x,y)}=x^2+(y+1)^2\\ maximize\quad d_{(x,y)}=x^2+(y+1)^2\\ \quad subject\ to,\\ \quad\quad x^2+16y^2=16\\$

Here objective function is $d_{(x,y)} =x^2+(y+1)^2$

and constraint is $C_{(x,y)}=x^2+16y^2-16=0$

so at the maximum point,

$\nabla d_{(x,y)}=\lambda*\nabla C_{(x,y)}\\\quad \lambda\ is\ the\ lagrange\ multiplier$

$\nabla d_{(x,y)}=\begin{bmatrix}2x \\2(y+1) \end{bmatrix}\\$

$\nabla C_{(x,y)}=\begin{bmatrix}2x \\32y\end{bmatrix}\\$

$\begin{bmatrix}2x \\2(y+1) \end{bmatrix}=\lambda\begin{bmatrix}2x \\32y\end{bmatrix}\\$

$2x=\lambda*2x\\ \lambda=1\\ then,\\ 2(y+1)=1*32y\\ y=\frac{1}{15}\\$

$using\ C_{(x,y)},\\ x^2+16*(\frac{1}{15})^2-16=0\\ x=\pm\frac{16}{15}\sqrt{14}\\$

Therefore furthest points are,

$$\blue{\bold{(\frac{16}{15}\sqrt{14},\frac{1}{15}) }}\ and\ \blue{ \bold{ (-\frac{16}{15}\sqrt{14},\frac{1}{15})}}\\$$

$d_{max}=(\frac{16}{15}\sqrt{14})^2+(\frac{1}{15}+1)^2=\frac{256}{45}$