Question 1209170: Find one ordered pair (x,y) of real numbers such that x + y = 8 and x^3 + y^3 = 200 - 4x^2 - 4y^2.
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
To help me get started on this, I first graphed the system of equations using desmos.com.
That graph showed that there are no real number solutions, so it would be a waste of time trying to solve the system algebraically.
ANSWER: no real solutions
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
Find one ordered pair (x,y) of real numbers such that x + y = 8 and x^3 + y^3 = 200 - 4x^2 - 4y^2.
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Left side x^3 + y^3 can be composed as (x+y)*(x^2 - xy + y^2).
The factor (x+y) can be replaced by 8, based on the first equation.
After that, second equation can be written in this form
8(x^2 - xy + y^2) = 200 - 4(x^2 + y^2),
or
12(x^2 + y^2) - 8xy = 200,
12(x+y)^2 - 24xy - 8xy = 200,
12(x+y)^2 - 32xy = 200.
Again, we can replace here (x+y) by 8, and we get then
12*8^2 - 32xy = 200,
32xy = 12*64 - 200 = 568,
xy = 568/32 = 17.75.
But under the condition x+y = 8, well known minimax property says
that the product xy may have the maximum value of 4*4 = 16 at x = y = 8/2 = 4.
In other words, under given conditions, the original system has no real solutions.
So, there is no any pair of real numbers (x,y) satisfying the imposed conditions.
Solved completely in Math terms.
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