Question

The sum of two nonnegative numbers is 20. Find the numbers if the sum of their squares is as large as possible; as small as possible

Answer 1

\(\color{#000000 }{ \displaystyle a+b=20 }\)
\(\color{#000000 }{ \displaystyle f(a,b)=a^2+b^2 }\)

Answer 2

maximize the function and minimize it

Answer 3

\(\color{#000000 }{ \displaystyle f(b)=(20-b)^2+b^2 }\)

Answer 4

is this calculus, btw, or not?

Answer 5

yes, in the section called applications of derivatives

Answer 6

I reduce the function to one variable...

Answer 7

but, for conveience, lets say rename the variables

\(\color{#000000 }{ \displaystyle x+w=20 }\)
\(\color{#000000 }{ \displaystyle f(x,w)=x^2+w^2 }\)

reducing to one variable
\(\color{#000000 }{ \displaystyle f(x)=x^2+(20-x)^2 }\)

Answer 8

oh, I got to go, sorrty

Answer 9

tag someone...

Answer 10

no probs ok

Answer 11

(I'm back online now....apologize for the delay)
I will do another kind of problem - similar one to this.
\(\color{#000000 }{ \huge \displaystyle ^\text{_____________________________} }\)
The sum of two nonnegative (real) numbers is 40. Find
the numbers if the sum of their cubes is (a) as large as
possible; (b) as small as possible.
\(\color{#000000 }{ \huge \displaystyle ^\text{_____________________________} }\)
So, let the numbers be \(\color{#0000ff }{ \displaystyle x }\) and \(\color{#0000ff }{ \displaystyle z }\), and then we know that
the numbers satisfy the following two eq-s/funct-s;

\(\color{#0000ff }{ \displaystyle x+z=40 }\) (sum of numbers is \(\color{#0000ff }{ \displaystyle 40 }\))
\(\color{#0000ff }{ \displaystyle f(x,z)=x^3+z^3 }\) (sum of cubes to \(\small\left\{\right.\)max/min\(\small\left.\right\}\)-ize)

It is easier to maximize the function with one variable,
and since we know that \(\color{#0000ff }{ \displaystyle x=40-z }\), we can substitute
that right into the sum of cubes.
\(\color{#0000ff }{ \displaystyle f(z)=(40-z)^3+z^3 }\)
(I like \(\color{#0000ff }{ \displaystyle z }\) better, but you can sub \(\color{#0000ff }{ \displaystyle x }\) same way)
I am going to assume that you know the algebra of it...
\(\color{#0000ff }{ \displaystyle f(z)=-z^3+120z^2-4800z+64000+z^3 }\)
\(\color{#0000ff }{ \displaystyle f(z)=120z^2-4800z+64000 }\)

\(\color{red}{■}\) Critical Number Theory
There are three cases when critical numbers occur for
a function \(f(x)\).
\(*1\) x-solutions for \(f'(x)=0\) (when the slope =0)
\(*2\) x-values for which \(f(x)\) is defined and \(f(x)\) isn't.
\(*3\) any closed boundaries (automatically).
\(\color{red}{\Downarrow}\) Continuation of the problem \(\color{red}{\Downarrow}\)

\(\color{#0000ff }{ \displaystyle f'(z)=240z-4800 }\)
NOTE: Case \(*2\) is excluded (a) you can clearly see that,
(b) f(x) is a polynomial.
\(\color{#0000ff }{ \displaystyle 0=240z-4800 }\)
\(\color{#0000ff }{ \displaystyle z =20}\)

The above result, \(\color{#0000ff }{ \displaystyle z =20}\), is from the Case \(*1\).
We also have \(\color{#0000ff }{ \displaystyle z =0}\) and \(\color{#0000ff }{ \displaystyle z =40}\) from the
Case \(*3\) because the sum of cubes, given by
\(\color{#0000ff }{ \displaystyle f(z)=(40-z)^3+z^3 }\), must satisfy \(\color{#0000ff }{ \displaystyle 0\le\{z,x\}\le40 }\)
(also note, that if one of the numbers is greater than
40, then the other is negative - and that can't be).

Now, let's plug the ciritcal numbers into the function.
\(\color{#0000ff }{ \displaystyle z=0 }\) \(\color{#0000ff }{ \displaystyle f(0)=(40-0)^3+0^3=40^3 }\)
\(\color{#0000ff }{ \displaystyle z=40 }\) \(\color{#0000ff }{ \displaystyle f(40)=(40-40)^3+40^3=40^3 }\)
\(\color{#0000ff }{ \displaystyle z=20 }\) \(\color{#0000ff }{ \displaystyle f(z)=(40-20)^3+20^3=2(20)^3 }\)
(And clearly, \(\color{#0000ff }{ \displaystyle 2(20)^3\le 40^3 }\))

The maximum sum of cubes is reached when \(\color{#0000ff }{ \displaystyle z=20 }\).
The minimum sum of cubes is reached when \(\color{#0000ff }{ \displaystyle z\in \{0,40\} }\).

we can repeat the answer when we include the second
number x, since we know that \(\color{#0000ff }{ \displaystyle x+z=40 }\).

ANSWER
The minimum sum of cubes is reached when \(\color{#0000ff }{ \displaystyle x,z =20 }\).
The maximum sum of cubes is reached when \(\color{#0000ff }{ \displaystyle (x,z)=(0,40)}\)

NOTE
Order doesn't matter for the maximum, since we found the
two numbers - 0 and 40, that maximize the sum of cubes.

Answer 12

EXAMPLE 2:

The sum of two nonnegative (real) numbers is 64. Find
the numbers if the sum of their perfect 5th's is;
Part 1) as large as possible
Part 2) as small as possible.
\(\color{#000000 }{ \huge \displaystyle ^\text{_____________________________} }\)
The sum of perfect 5th's is written as:
\(\color{#000000 }{ \displaystyle x+z=64 }\)
The function we want to maximize is,
\(\color{#000000 }{ \displaystyle f(x,z)=z^5+x^5;\quad\quad 0\le \{x,z\}\le64 }\)
(The boundaries for \(\color{#000000 }{ z }\) and \(\color{#000000 }{ x }\) come from the first equation,
and from the condition that \(\color{#000000 }{ z }\) and \(\color{#000000 }{ x }\) are nonnegatives.)

\(\color{#000000 }{ \displaystyle f(x)=(64-x)^5+x^5\quad\quad 0\le x\le64 }\)
(I used \(x\) for some diversity; could've used \(z\) as well,
and note that the boundary for \(x\) is maintained)

Expanding is painful, so let's differentiate it as it is:
\(\color{#000000 }{ \displaystyle f'(x)=5(64-x)^4\times (-1)+5x^4 }\)

\(x=0,64\) are critical numbers automatically \(\small (\)via Case \(*3\)
from aforementioned "Critical Number Theory"\(\small )\)
Therefore only Case \(*1\) remains to work on.

\(\color{#000000 }{ \displaystyle 0=5(64-x)^4\times (-1)+5x^4 }\)
\(\color{#000000 }{ \displaystyle 5(64-x)^4=5x^4 }\)
\(\color{#000000 }{ \displaystyle (64-x)^4=x^4 }\)
\(\color{#000000 }{ \displaystyle \sqrt[4]{(64-x)^4}=\sqrt[4]{x^4} }\)

The absolute value of \(\varphi\) (I used that symbol, greek,
doesn't matter) is given as, \(\left|\varphi \right|=\sqrt{\varphi^2}\), and when
you have something like \(\sqrt[4]{\varphi^4}\), then this is
(certainly/also) an absolute value.
\(\color{#000000 }{ \displaystyle \left|64-x\right|= \left|x\right| }\)
We have two cases, but only one is applicable,
\(\color{#000000 }{ \displaystyle 64-x=x }\) \(\color{#000000 }{ \displaystyle 64-x=-x }\)
\(\color{#000000 }{ \displaystyle 64=2x }\) \(\color{#000000 }{ \displaystyle 64=0 }\) (oops ... error\(!\))
\(\color{#000000 }{ \displaystyle x=32 }\) Unapplicable\(/\)False

So, \(\color{#000000 }{ \displaystyle x=32 }\)

Overall, the critical numbers we obtain are:
\(\color{#000000 }{ \displaystyle x=0,32,64 }\)

recall that,
\(\color{#000000 }{ \displaystyle f(x)=(64-x)^5+x^5 }\)

let's do the plugging,
\(\color{#000000 }{ \displaystyle f(0)=(64-0)^5+0^5=64^5 }\)
\(\color{#000000 }{ \displaystyle f(64)=(64-64)^5+64^5=64^5 }\)
\(\color{#000000 }{ \displaystyle f(8)=(64-32)^5+32^5=2(32)^5 }\)
(hope we know that \(\color{#000000 }{ \displaystyle 2(32)^5<64^5 }\))

Via \(\color{#000000 }{ \displaystyle x+z=64 }\);
When x=0,64, then z=64,0, respectively.
When x=32, then z=32.

So, the ANSWER is:

▼ Maximum sum of perfect 5th's is reached when \( (x,z)=(0,64) \).
▼ Minimum sum of perfect 5th's is reached when \( (x,z)=(32,32) \).

(Again the order of 0 and 64, or the order of 32 and 32, doesn't matter).

Answer 13

In fact, you can prove for any perfect \(n\)th's, if
\(\color{#000000 }{ \displaystyle x+z=\beta }\) and \(\color{#000000 }{ \displaystyle 0\le x,z\le \beta }\), that ...

■ maximum will be \(\color{#000000 }{ \displaystyle \beta^n }\) and it will occur at \(\color{#000000 }{ \displaystyle (x,z)=(0,\beta) }\)
■ minimum will be \({\displaystyle\color{black}{2\left(\frac{\beta}{2}\right)^n}}\) and it will occur at \(\color{#000000 }{ \displaystyle (x,z)=\left(\frac{\beta}{2},\frac{\beta}{2}\right) }\)

((Note that you set \(\color{#000000 }{ \displaystyle f(x,z)=x^n+z^n }\)))

Answer 14

THE PROOF\( : \)

GIVEN/SETUP\( : \)
\(\color{#000000 }{ \displaystyle x+z=\beta }\) (sum of 2 nonnegatives which is \(=\beta\))
\(\color{#000000 }{ \displaystyle f(x,z)=x^n+z^n }\) (SUM OF PERFECT \(n\)th's)

CONSTRAINT/CONDITIONS\( : \)
\(\color{#000000 }{ \displaystyle \beta\in\mathbb{R},~\beta>0 }\)
\(\color{#000000 }{ \displaystyle n\in\mathbb{N},~n\ge 2 }\)
\(\color{#000000 }{ \displaystyle 0\le \{x,z\} \le \beta }\)


SUBSTITUTION\( : \)
\(\color{#000000 }{ \displaystyle x+z=\beta\quad \Longrightarrow \quad x=\beta -z}\)
\(\color{#000000 }{ \displaystyle f(z)=(\beta -z)^n+z^n }\)


DERIVATIVE/F'(Z)=0\( : \)
\(\color{#000000 }{ \displaystyle f'(z)=-n(\beta -z)^{n-1}+nz^{n-1} }\)
\(\color{#000000 }{ \displaystyle 0=-n(\beta -z)^{n-1}+nz^{n-1} }\)
\(\color{#000000 }{ \displaystyle n(\beta -z)^{n-1}=nz^{n-1} }\)
\(\color{#000000 }{ \displaystyle (\beta -z)^{n-1}=z^{n-1} }\)
\(\color{#000000 }{ \displaystyle \sqrt[n-1]{(\beta -z)^{n-1}}=\sqrt[n-1]{z^{n-1}} }\)


CASE 1\( : \) (n-1) is even
\(\color{#000000 }{ \displaystyle |\beta-z|=|z| }\)
\(\color{#000000 }{ \displaystyle \beta-z=z}\) \(\color{#000000 }{ \displaystyle \beta-z=-z}\)
\(\color{#000000 }{ \displaystyle \beta=2z}\) \(\color{#000000 }{ \displaystyle \beta=0}\)
\(\color{#000000 }{ \displaystyle x=\beta/2}\) (not a solution)


CASE 2\( : \) (n-1) is odd
\(\color{#000000 }{ \displaystyle \beta-z=z\quad \Longrightarrow\quad z=\beta/2 }\) \(\small (\)Same as CASE 1\(\small )\)


CRITICAL NUMBERS\( : \)
\(\color{#000000 }{ \displaystyle z=0}\) \(\small (\)Closed Interval Boundary\(\small )\)
\(\color{#000000 }{ \displaystyle z=\beta/2}\) \(\small (\)Solution to \(f'(x)=0\)\(\small )\)
\(\color{#000000 }{ \displaystyle z=\beta}\) \(\small (\)Closed Interval Boundary\(\small )\)


PLUGGING FOR MAX/MIN\( : \)
\(\color{#000000 }{ \displaystyle f(0)=f(\beta)=\beta^n }\)
\(\color{#000000 }{ \displaystyle f\left(\beta/2\right) =2\left(\beta/2\right)^n}\)

WHICH IS MAX AND WHICH IS MIN(?)
(Recall the "CONDITIONS" previously listed)
\(\color{#000000 }{ \displaystyle 2\left(\beta/2\right)^n }\) \(\color{#000000}{\square}\) \(\color{#000000 }{ \displaystyle \beta^n }\)
\(\color{#000000 }{ \displaystyle 2\beta^n/2^{-n} }\) \(\color{#000000}{\square}\) \(\color{#000000 }{ \displaystyle \beta^n }\)
\(\color{#000000 }{ \displaystyle \beta^n/2^{1-n} }\) \(\color{#000000}{\square}\) \(\color{#000000 }{ \displaystyle \beta^n }\)
\(\color{#000000 }{ \displaystyle 1/2^{1-n} }\) \(\color{#000000}{\square}\) \(\color{#000000 }{ \displaystyle 1 }\)
\(\color{#000000 }{ \displaystyle 1 }\) \(\color{#000000}{\square}\) \(\color{#000000 }{ \displaystyle 2^{n-1} }\)
\(\color{#000000 }{ \displaystyle 1 }\) \({\color{magenta}{\bf <}}\) \(\color{#000000 }{ \displaystyle 2^{n-1} }\)

Knowing that;
So, \(\color{black}{\forall~n\in \mathbb{N}}\), and \(\color{black}{n>1}\) we can conclude,
\(\color{#000000 }{ \displaystyle 2\left(\beta/2\right)^n }\) \({\color{magenta}{\bf <}}\) \(\color{#000000 }{ \displaystyle \beta^n }\)
(MIN) (MAX)

FINDING THE OTHER NUMBER\( : \)
\(\color{#000000 }{ \displaystyle x=\beta-z }\)
\(\color{#000000 }{ \displaystyle z=\beta/2 }\) \(\color{#000000 }{ \displaystyle x=\beta-\beta/2=\beta/2}\)
(So x and z, are both \(\color{#000000 }{ \displaystyle \beta/2 }\))

\(\color{#000000 }{ \displaystyle z=\beta }\) \(\color{#000000 }{ \displaystyle x=\beta-\beta=0 }\)
\(\color{#000000 }{ \displaystyle z=0 }\) \(\color{#000000 }{ \displaystyle x=\beta-0=\beta }\)
(So x and z, are \(\color{#000000 }{ \displaystyle 0 }\) and \(\color{#000000 }{ \displaystyle \beta }\); order doesn't matter)


CONCLUSION\( : \)
MAXIMUM sum of 2 perfect \(n\)th's is \(\color{#000000 }{ \displaystyle \beta^n }\).
Occurs when the numbers are \(\color{#000000 }{ \displaystyle 0 }\) and \(\color{#000000 }{ \displaystyle \beta }\).
MINIMUM sum of 2 perfect \(n\)th's is \(\color{#000000 }{ \displaystyle 2(\beta/2)^n }\).
Occurs when the numbers are both \(\color{#000000 }{ \displaystyle \beta/2 }\).


PROVED \( !! \)