## Brent0423 3 years ago Find two positive numbers whose sum is 100 and the sum of whose squares is a minimum.

1. Brent0423

Is it just me or is this problem pretty complex?

2. Brent0423

this is what all of my problems are like... FML

3. ChmE

I don't understand squares is a minimum... hmmm

4. Algebraic!

x+y=100 x^2+y^2= something x^2 + (100-x)^2 = something differentiate set equal to zero

5. Brent0423

i know how to do the problem now.. i just feel like these problems are really complex

x+y=100 ----> y=100-x k=x^2+y^2 k=x^2+(100-x)^2 k=x^2+10,000-200x+x^2 k=2x^2-200x+10,000 u can use derivative like @Algebraic! said

7. Brent0423

A rancher with 750ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. a. Find a function that models the total area of the four pens. b. Find the largest possible total area of the four pens.

the 2nd problem ?

9. Brent0423

ya......

|dw:1359869626673:dw| from the figure we can maka a models the total area of the four pens : 6x+4y=750 or 3x+2y=375 2y=375-3x y=(375-3x)/2 ...(1) the total area (A) = 4xy ....(2) subt (1) to (2) we get a function : A=4x(375-3x)/2 A=-6x^2+750x

11. Brent0423

and your trying to tell me that this is easy....... -_-

ups, sorry mistake from m the total area (A) = 3xy A=3x(375-3x)/2 A=562.5x-4.5x^2

it just a model only :) for solution u can use like the first problem

btw, the first has solved ?

15. akash_809

the correct way is to diffentiate and equate it to zero ...But i think there is an easier method , since $(x+y)^2=x^2+y^2+2xy$ so $10000=x^2+y^2+2xy$ now from A.M $\ge$ GM , so $(x^2+y^2)/2 \ge \sqrt{x^2.y^2}$ and for $x^2+y^2$ to be minimum A.M=G.M i.e x=y , so the numbers would be 50,50 ....

16. akash_809

@Brent0423 do u get it

17. akash_809

and do you have theanswer , is it 50,50 i gave

18. akash_809

you gave me the medal so i assume the answer is 50, 50 right ?

19. Brent0423

yes