**#Problem 9**

a+b+c+d+e=8

a^2+b^2+c^2+d^2+e^2=16. [a,b,c,d,e are real numbers]

**Solution Scheme and approach:**

This problem can easily be done with Cauchy-Schwartz inequality. But will explain this in my next topic. Let's do it without Cauchy.

(a-r)^2+(b-r)^2+(c-r)^2+(d-r)^2+(e-r)^2

=(a^2+b^2+c^2+d^2+e^2)-2r(a+b+c+d+e)+5r^2

=16-16r+5r^2[ By putting the values]

we can write

(e-r)^2<=16-16r+5r^2

=>(e-r)=root[16-16r+5r^2] [equality holds when a=b=c=d=r]

=>e=root[16-16r+5r^2]+r=f(r)

as this function is increasing we have only local minima

f'(r)=0

=>(10r-16)/root(16-16r+5r^2)=0...(i)

=>r=2 or 6/5

but r=2 is extraneous solution.Does not satisfy eqn number (i)

So r=6/5

=>e is maximum when a=b=c=d=6/5

=>max(e)=f(6/5)=16/5.

16/5 is the ans by cauchy--

ReplyDeleteWell done :)

ReplyDelete