Suppose each of x and y is a real number > 0
Find the minimum value of:
2x2+2y2+3xy+1
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x+y
Let V = the expression.
On inspection, x and y are interchangeable.
What if x = y?
If x=y , V = (7x^2 + 1)/2x
V' = (28x^2 - 14x^2 - 2)/4x^2
V' = (14x^2 - 2)/4x^2 = 7/2 - 1/2x^2 = 0
x^2 = 1/7, and x>0 so x=y=√7/7
And then the expression equals: (7/7 + 1)/(2/√7) = √7.
Let a = √7/7, then calculate f(a,a) and f(a+epsilon, a-epsilon), where epsilon is a very small number.
It turns out the latter is slightly larger, further evidence that the minimum has been found.
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Posted by Larry
on 2024-05-18 09:24:55 |
Possible solution
