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Prove that if you minimize the square of the distance from the origin to a point (x, y) subject to the constraint g(x,y)=0, you have minimized the distance from the origin to (x, y) subject to the same constraint.

Short Answer

Expert verified

d=x2+y2,∇d(x,y)=xx2+y2i+yx2+y2j∇g(x,y)=0.Solvingwegetminimumat(0,0).D=x2+y2,∇D(x,y)=(2x)i+(2y)j∇g(x,y)=0.Solvingwegetminimumat(0,0).Therefore,inbothcaseswegetthesamesolution.

Step by step solution

01

Step 1. Given Information.

Given the constraint: g(x,y)=0.

And the point (x, y) is given on the plane.

02

Step 2. Minimize the distance from origin to (x, y).

Let d be the distance from origin to the point.

d=x2+y2.

∇d(x,y)=∂d∂xi+∂d∂yj∇d(x,y)=xx2+y2i+yx2+y2jAndgiven,∇g(x,y)=0.

So, by Lagrange's method:

localid="1649910485059" ∇d=λ∇g,xx2+y2i+yx2+y2j=λ0i+0j,Thisimplies,xx2+y2=0andyx2+y2=0Whichisequaltox=0andy=0.Therefore,theminimumoccursat(0,0).

03

Step 2. Minimize the square of the distance from origin to (x, y).

Let D be the distance from origin to the point.

D=x2+y2.∇D(x,y)=∂D∂xi+∂D∂yj∇D(x,y)=2xi+2yjAndgiven,∇g(x,y)=0.

So, by Lagrange's method:

∇D=λ∇g,2xi+2yj=λ0i+0j,Thisimplies,2x=0and2y=0Whichisequaltox=0andy=0.Therefore,theminimumoccursat(0,0).

Therefore, in both the cases the solution is the same.

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