Answer to Question #107937 in Calculus for annie

Question #107937
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.
1
Expert's answer
2020-04-14T15:05:46-0400

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


problem is,


d(x,y)=(x0)2+(y(1))2d(x,y)=x2+(y+1)2maximized(x,y)=x2+(y+1)2subject to,x2+16y2=16d_{(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)=x2+(y+1)2d_{(x,y)} =x^2+(y+1)^2

and constraint is C(x,y)=x2+16y216=0C_{(x,y)}=x^2+16y^2-16=0

so at the maximum point,

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


d(x,y)=[2x2(y+1)]\nabla d_{(x,y)}=\begin{bmatrix}2x \\2(y+1) \end{bmatrix}\\

C(x,y)=[2x32y]\nabla C_{(x,y)}=\begin{bmatrix}2x \\32y\end{bmatrix}\\

[2x2(y+1)]=λ[2x32y]\begin{bmatrix}2x \\2(y+1) \end{bmatrix}=\lambda\begin{bmatrix}2x \\32y\end{bmatrix}\\

2x=λ2xλ=1then,2(y+1)=132yy=1152x=\lambda*2x\\ \lambda=1\\ then,\\ 2(y+1)=1*32y\\ y=\frac{1}{15}\\

using C(x,y),x2+16(115)216=0x=±161514using\ C_{(x,y)},\\ x^2+16*(\frac{1}{15})^2-16=0\\ x=\pm\frac{16}{15}\sqrt{14}\\

Therefore furthest points are,


(161514,115) and (161514,115)\blue{\bold{(\frac{16}{15}\sqrt{14},\frac{1}{15}) }}\ and\ \blue{ \bold{ (-\frac{16}{15}\sqrt{14},\frac{1}{15})}}\\



dmax=(161514)2+(115+1)2=25645d_{max}=(\frac{16}{15}\sqrt{14})^2+(\frac{1}{15}+1)^2=\frac{256}{45}


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