Find two positive real numbers whose sum is 100 and whose product is a maximum?

- anonymous

Find two positive real numbers whose sum is 100 and whose product is a maximum?

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- whpalmer4

You can think of this as \[y = x*(100-x)\] which is an inverted parabola. What is the vertex?

- anonymous

Domain is (100, infinity)?

- ash2326

@kissy Let the numbers be x and y
we have a condition
\[x+y=100\]
Do you get this part?

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## More answers

- anonymous

absolutely

- ash2326

Have you studies calculus?

- ash2326

*studied

- anonymous

no, sorry. That's why i'm on here ):

- ash2326

I just want to know if you know differentiation and how to find the maxima and minima for a function. Of course I'll explain

- anonymous

Oh yes, I understand how to differentiate and how to find the extrema. I'm just having trouble understanding how to solve optimization problems

- ash2326

Ok, let's work on it

- ash2326

Let the product of x and y be P
so
\[P= x\times y\]
We know y=100-x
so
\[P=x\times (100-x)\]
Do you get this part?

- anonymous

yes

- ash2326

So we want to maximize P
Let's differentiate P with respect to x. Can you do that?

- anonymous

use chain rule to differentiate, right?

- anonymous

so P'(x)= 1(100-x)*(1)

- ash2326

It should be
\[P'(x)=1 \times (100-x)+x\times (0-1)\]

- anonymous

Shoot, forgot the x. Sorry

- ash2326

No problem, so we get
\[P'(x)=100-x-x\]
Can you find value of x is P'(x)=0

- anonymous

10?

- ash2326

\[0=100-x-x\]
\[0=100-2x\]
Can you try now?

- anonymous

Oh, I thought you meant what makes P'(x) 0 but you meant make P'(x) equal to 0. It is 50

- ash2326

Cool. I understood that. So there you go you got x , can you find y?

- anonymous

To find y you would make P'(0) right? ... as in you plug 0 into the x's of the original formula

- ash2326

We have the original condition
\[x+y=100\]
put x=50 and find y from this.

- anonymous

50 then.... where did you get x+y=100?

- anonymous

...nevermind. The questions states that the sum of x and y is 100 lol

- ash2326

Cool, so these are the two no.s which will make product the maximum

- anonymous

they are the CN's?

- ash2326

CN?

- anonymous

critical numbers

- ash2326

yes

- ash2326

do you understand?

- anonymous

Yes I do. I'm still a bit confused still though... like I understand what you did and i'm just still not understanding the rest of the steps

- ash2326

Just read it again, all of my posts. Let me know wherever you have doubt

- anonymous

Alright. I will look through again and try to figure it out on my own but if I still don't understand it I will let you know (: thanks

- ash2326

Cool

- anonymous

no calculus needed
the sum is 100, the product is a max, it is \(50\times 50\)

- anonymous

I still don't quite understand

- anonymous

ok you have two numbers whose sum is 100 right?

- anonymous

yes

- anonymous

so for example one could be 30 and the other 70
or one could be 25 and the other 75
or one could be 99 and the other 1
or even one could be 100 and the other one 0
in other words, you have two number \(x\) and \(y\) where \(x+y=100\) or if you prefer you have two numbers \(x\) and \(100-x\)

- anonymous

Oh, so the x and y that we just found are the two positive real numbers

- anonymous

yeah, just some real numbers
now here is the thing: you cannot tell then apart
by which i mean if you say one is \(x\) so the other is \(y=100-x\) or i say one is \(100-x\) and the other is \(x\) we both have the same thing, two numbers whose sum is 100

- anonymous

since you cannot distinguish between the numbers, \(x\) and \(y\) or \(x\) and \(100-x\), i.e. since they are interchangeable, the procedure for finding the maximum product cannot favor one number over the other
you say \(x+y=100\) and i say \(y+x=100\) we are both saying the same thing

- anonymous

so the maximum must be when the two numbers are identical, because we cannot tell them apart

- anonymous

you can check it with numbers and see what i mean
you say the first number is \(30\) so the second number must be \(70\) and \(30\times 70=2100\) now if i interchange them, we get \(70\times 30=2100\) the same answer exactly
now they will be biggest if we make both numbers the same, i.e. if \(x=y=50\) and \(50\times 50=2500\) is the biggest

- anonymous

Ok, I understand. So it'd be 50x50 for max and the two positive real numbers are 50 and 50 because like you said about the identical thing

- anonymous

if you don't like my logic which requires only common sense, we can also do it using algebra

- anonymous

right, you cannot tell them apart, so this procedure of finding the max cannot favor one number over the other
but we can still use algebra if this will make your teacher happy

- anonymous

Alright, how would you do it the algebra way? (it's not required but i'd still like to understand it)

- anonymous

call one number \(x\) so the other number must be \(100-x\) and their product is \(x(100-x)=100x-x^2\)

- anonymous

the equation
\[y=100x-x^2\] is a parabola that opens down
the biggest it can be is at the vertex, and the first coordinate of the vertex of a parabola is \(-\frac{b}{2a}\) which in this case is \(-\frac{100}{2\times (-1)}=50\)

- anonymous

therefore the maximum the product can be is if \(x=50\) which of course means the other number is \(50\) as well, and \(50\times 50=2500\)

- anonymous

it is simple enough algebra, but common sense is even simpler
how could it be possible that the maximum of the product could anything other than the number you would get if the two numbers were equal? why would one number be favored over the other? in other words, since \(x+y=100\) is symmetric in \(x\) and \(y\) by which i mean \(x+y=100\) is exactly the same as \(y+x=100\) it must be the case that the max is if \(x=y\)

- anonymous

Thank you!

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