Given that x and y vary so that ax + by= c , where a,b,c are constant. Show that the minimum value of x^2 + y^2 is c^2/ a^2 + b^2 ?

2 Answers
Feb 22, 2017

The minimum value is c^2/(a^2 + b^2)

Explanation:

The minimum of x^2+y^2 restricted to ax +by=c is obtained by minimizing

f(x)=x^2+((c-ax)/b)^2 The condition of relative minimum is

(df)/(dx)=0 at x = x_0 and

(d^2f)/(dx^2) > 0 at x=x_0

Then

(df)/(dx)=2x-2(a/b)((c-ax)/b)=0-> x_0 = (a c)/(a^2+b^2)

substituting this value into f(x) we have

f(x_0)=c^2/(a^2 + b^2)

Also

(d^2f)/(dx^2)=2 + (2 a^2)/b^2 > 0

Feb 23, 2017

(x^2+y^2)_min=c^2/(a^2+b^2).

Explanation:

Let O(0,0) be the Origin, and, (x,y) be the General

Point in the Plane RR^2.

If L={(x,y) : ax+by=c; a,b,c in RR}, then, we know that, L is

a Family of Lines in the plane, where, a^2+b^2ne0.

Let us note that, c=0 iff O(0,0) in L.

Now, the Distance OP=sqrt(x^2+y^2) rArr x^2+y^2=OP^2.

Accordingly, when (x^2+y^2) is Minimum, so is the Distance

OP.

Thus, in this new scenario, our Problem is to find the

Minimum Distance from the Origin O(0,0)" to the Line "L.

From Geometry, we know that the minimum distance of a point

(x',y') to a line : Ax+by+C=0 is the bot"-distance" between

them, given by, |Ax'+by'+C|/sqrt(A^2+b^2).

Accordingly, {sqrt(x^2+y^2)}_(min)=|c|/sqrt(a^2+b^2), or,

(x^2+y^2)_min=c^2/(a^2+b^2)......................(star)

We conclude with a note that, in case, O(0,0) in L, i.e., c=0,

(x^2+y^2)_min=0, Geometrically, and, in accordance with

(star) as well.

Enjoy Maths.!