How to choose two numbers for which the sum of their square roots is minimal, knowing that the product of the two numbers is #a#?

How can I solve the problem with derivatives?

3 Answers
Feb 14, 2018

#x=y=sqrt(a)#

Explanation:

#x*y=a => x*y - a = 0#
#f(x,y) = sqrt(x)+sqrt(y) " is minimal"#
#"We could work with the Lagrange multiplier L :"#
#f(x,y,L) = sqrt(x)+sqrt(y)+L(x*y-a)#

#"Deriving yields :"#

#{df}/dx = 1/(2*sqrt(x)) + L*y = 0#
#{df}/dy = 1/(2*sqrt(y)) + L*x = 0#
#{df}/{dL} = x*y-a = 0#

#=> y=a/x#
#=> {df}/dy = 1/(2*sqrt(a/x)) + L*x = 0#
#= sqrt(x)/(2*sqrt(a)) + L*x = 0#
#=> {df}/dx = 1/(2*sqrt(x)) + L*a/x = 0#
#=> sqrt(x)/2 + L*a = 0 " (after multiplying with x"!="0)"#
#=> L = - sqrt(x)/(2*a)#
#=> sqrt(x)/(2*sqrt(a)) - sqrt(x)*x/(2*a) = 0#
#=> 1/(2*sqrt(a)) - x/(2*a) = 0#
#=> x = sqrt(a)#
#=> y = sqrt(a)#
#=> L = -a^(1/4)/(2*a) < 0 => " MINIMUM"#

#"Now we still have to check x=0."#
#"This is impossible as x*y = 0 then."#

#"So we have the unique solution"#
#x=y=sqrt(a)#

Feb 14, 2018

I'll try to take you through the solution method below.

Explanation:

What are we looking for?

Two numbers. Let's give them names, #x# and #y#.

Reread the question.

We want to make the sum of the square roots minimal.

This tells us two things
(1) both numbers are non-negative (to avoid imaginaries)
(2) We are interested in the value of #sqrtx+sqrty#

Reread the question.

We also are told that the product of #x# and #y# is #a#.

Who chooses #a#?

In general, if an exercise says something about #a# or #b# or #c#, we take those as constants given by someone else.

So we might be told "the product of #x# and #y# is #11#"
or "the product of #x# and #y# is #124#".

We are to solve all of these at once by saying #xy=a# for some constant #a#.

So, we want to make #sqrtx+sqrty# as small as possible keeping #xy=a# for some constant #a#.

This looks like an optimization problem and it is one. So I want a function of one variable to minimize.
#sqrtx+sqrty# has two variables, #x# and #y#

#xy=a# also has two variables, #x# and #y# (remember #a# is a constant)

So #y = a/x#

Now we want to minimize:

#f(x) = sqrtx+sqrt(a/x) = sqrtx+sqrta/sqrtx#

Find the derivative, then the critical number(s) and test the critical number(s). Finish be finding #y#.

#f'(x) = (x-sqrta)/(2x^(3/2))#

Critical #sqrta#

#f'(x) < 0# for #x < sqrta# and #f'(x) > 0# for #x > sqrta#, so #f(sqrta)# is a minimum.

#x = sqrta# and #y = a/x = sqrta#

Feb 14, 2018

#2 root(4)(a)#

Explanation:

We know that for #x_i > 0# we have

#(x_1 x_2 cdots x_n)^{\frac{1}{n}} \le \frac{x_1+ x_2 +cdots + x_n}{n}#

then

#x_1+x_2 ge 2 sqrt(x_1 x_2)# then

#sqrtx_1 + sqrt x_2 ge 2 root(4)(x_1x_2)#

but #x_1x_2 = a# then

#sqrtx_1 + sqrt x_2 ge 2 root(4)(a)#