SOLUTION: If x and y are two positive real nos. such that their sum is 1, then find the maximum value of {{{xy^4 + yx^4}}} (Please solve using basic algebra and logic and NOT using calculus

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: If x and y are two positive real nos. such that their sum is 1, then find the maximum value of {{{xy^4 + yx^4}}} (Please solve using basic algebra and logic and NOT using calculus      Log On


   

Question 977299: If x and y are two positive real nos. such that their sum is 1, then find the maximum value of xy%5E4+%2B+yx%5E4
(Please solve using basic algebra and logic and NOT using calculus)
Found 2 solutions by josgarithmetic, Edwin McCravy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
x+y=1
y=1-x
-
x%281-x%29%5E4%2B%281-x%29x%5E4
x%281-x%29%28%281-x%29%5E3%2Bx%5E3%29
x%281-x%29%28-x%5E3-x%5E2-3x%2B1%2Bx%5E2%29
x%281-x%29%28-x%5E2-3x%2B1%29
%28-1%29x%281-x%29%28x%5E2%2B3x-1%29

You may be able to identify roots of the factored expression and on which intervals is the expression positive. You could check using iterations of numerical values to find max values as near as you want for any maximum values which exist.

The ROOTS (not any extremes) are 0, 1, (-3-sqrt(13))/2, (-3+sqrt(13))/2.
Remember you wanted positive values for x, and that 0%3Cx%3C1.

I have not worked through to any finished answer; just planned for the method.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
If x and y are two positive real nos. such that their sum is 1, then find the maximum value of xy%5E4+%2B+yx%5E4
(Please solve using basic algebra and logic and NOT using calculus)
letz=xy%5E4+%2B+yx%5E4
z=xy%28y%5E3%2Bx%5E3%29
z=xy%28y%2Bx%29%28y%5E2-xy%2Bx%5E2%29
z=xy%28y%2Bx%29%28y%5E2-xy%2Bx%5E2%29

Since x+y=1, substitute 1-x for y

z=x%281-x%29%28%281-x%29%2Bx%29%28%281-x%29%5E2-x%281-x%29%2Bx%5E2%29
z=x%281-x%29%281%29%281-2x%2Bx%5E2-x%2Bx%5E2%2Bx%5E2%29
z=x%281-x%29%281-3x%2B3x%5E2%29
z=%28x-x%5E2%29%281-3x%2B3x%5E2%29
z=x-3x%5E2%2B3x%5E3-x%5E2%2B3x%5E3-3x%5E4
z=-3x%5E4%2B6x%5E3-4x%5E2%2Bx

If we can find A, B and nonnegative C such that

z=-3%28x%5E2%2BAx%2BB%29%5E2%2BC

then it's easy to see that z will have its maximum value 
when the quadratic x%5E2%2BAx%2BB equals 0, and that maximum
value will be C. 

Let's see if we can find such A, B, and C by equating
coefficients:

-3x%5E4%2B6x%5E3-4x%5E2%2Bx+=+-3%28x%5E2%2BAx%2BB%29%5E2%2BC

-3x%5E4%2B6x%5E3-4x%5E2%2Bx+=+-3x%5E4-3A%5E2x%5E2-3B%5E2-6Ax%5E3-6Bx%5E2-6ABx%2BC

Equating coefficients of x³:  6+=+-6A, or A+=+-1

Equating coefficients of x²:  -4+=+-3A%5E2-6B
                              -4+=+-3%28-1%29%5E2-6B
                              -4+=+-3%281%29-6B
                              -4+=+-3-6B
                              -1+=+-6B
                              1%2F6+=+B

Equating coefficients of x:  1+=+-6AB
                             1+=+-6%28-1%29%281%2F6%29
                             1+=+1  

                              Already met.


Equating constant terms:      0+=+-3B%5E2%2BC
                              0+=+-3%281%2F6%29%5E2%2BC
                              0+=+-3%281%2F36%29%2BC
                              0+=+-3%2F36%2BC
                              3%2F36=C
                              1%2F12=C

That's the answer C=1%2F12

z will take on that maximum value 1/12 when the quadratic x%5E2%2BAx%2BB
takes on the value 0.  

x%5E2%2BAx%2BB+=+0
x%5E2-1x%2B1%2F6+=+0

x+=+%281+%2B-+sqrt%281-2%2F3%29%29%2F2+
x+=+%281+%2B-+sqrt%281%2F3%29%29%2F2+
x+=+%281+%2B-+sqrt%283%29%2F3%29%2F2+
Multiply top and bottom by 3

x+=+%283+%2B-+sqrt%283%29%29%2F6

So there are two points where z reaches the maximum value of 1%2F12

They are %28+matrix%281%2C3%2C%283+%2B-+sqrt%283%29%29%2F6%2C%22%2C%22%2C1%2F12%29%29

Edwin