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# ?

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Answer 1 · 2017-02-22T18:03:24

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

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#

Answer 2 · 2017-02-23T02:11:42

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

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.!