anonymous
  • anonymous
Find two positive real numbers whose sum is 100 and whose product is a maximum?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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whpalmer4
  • whpalmer4
You can think of this as \[y = x*(100-x)\] which is an inverted parabola. What is the vertex?
anonymous
  • anonymous
Domain is (100, infinity)?
ash2326
  • 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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anonymous
  • anonymous
absolutely
ash2326
  • ash2326
Have you studies calculus?
ash2326
  • ash2326
*studied
anonymous
  • anonymous
no, sorry. That's why i'm on here ):
ash2326
  • 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
  • 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
  • ash2326
Ok, let's work on it
ash2326
  • 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
  • anonymous
yes
ash2326
  • ash2326
So we want to maximize P Let's differentiate P with respect to x. Can you do that?
anonymous
  • anonymous
use chain rule to differentiate, right?
anonymous
  • anonymous
so P'(x)= 1(100-x)*(1)
ash2326
  • ash2326
It should be \[P'(x)=1 \times (100-x)+x\times (0-1)\]
anonymous
  • anonymous
Shoot, forgot the x. Sorry
ash2326
  • ash2326
No problem, so we get \[P'(x)=100-x-x\] Can you find value of x is P'(x)=0
anonymous
  • anonymous
10?
ash2326
  • ash2326
\[0=100-x-x\] \[0=100-2x\] Can you try now?
anonymous
  • anonymous
Oh, I thought you meant what makes P'(x) 0 but you meant make P'(x) equal to 0. It is 50
ash2326
  • ash2326
Cool. I understood that. So there you go you got x , can you find y?
anonymous
  • anonymous
To find y you would make P'(0) right? ... as in you plug 0 into the x's of the original formula
ash2326
  • ash2326
We have the original condition \[x+y=100\] put x=50 and find y from this.
anonymous
  • anonymous
50 then.... where did you get x+y=100?
anonymous
  • anonymous
...nevermind. The questions states that the sum of x and y is 100 lol
ash2326
  • ash2326
Cool, so these are the two no.s which will make product the maximum
anonymous
  • anonymous
they are the CN's?
ash2326
  • ash2326
CN?
anonymous
  • anonymous
critical numbers
ash2326
  • ash2326
yes
ash2326
  • ash2326
do you understand?
anonymous
  • 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
  • ash2326
Just read it again, all of my posts. Let me know wherever you have doubt
anonymous
  • 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
  • ash2326
Cool
anonymous
  • anonymous
no calculus needed the sum is 100, the product is a max, it is \(50\times 50\)
anonymous
  • anonymous
I still don't quite understand
anonymous
  • anonymous
ok you have two numbers whose sum is 100 right?
anonymous
  • anonymous
yes
anonymous
  • 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
  • anonymous
Oh, so the x and y that we just found are the two positive real numbers
anonymous
  • 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
  • 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
  • anonymous
so the maximum must be when the two numbers are identical, because we cannot tell them apart
anonymous
  • 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
  • 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
  • anonymous
if you don't like my logic which requires only common sense, we can also do it using algebra
anonymous
  • 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
  • anonymous
Alright, how would you do it the algebra way? (it's not required but i'd still like to understand it)
anonymous
  • anonymous
call one number \(x\) so the other number must be \(100-x\) and their product is \(x(100-x)=100x-x^2\)
anonymous
  • 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
  • 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
  • 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
  • anonymous
Thank you!

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