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07 Jun 2019, 01:04
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D
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Difficulty:
35%
(medium)
Question Stats:
74% (01:46) correct
26%
(02:18)
wrong
based on 134
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The least value of \(x^2 + 5xy + 4y^2\) given x + y = 2 and x, y are non negative, is:
A. 0
B. 2
C. 3
D. 4
E. 6
A. 0
B. 2
C. 3
D. 4
E. 6
07 Jun 2019, 02:06
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nick1816
The least value of \(x^2 + 5xy + 4y^2\) given x + y = 2 and x, y are non negative, is:
A. 0
B. 2
C. 3
D. 4
E. 6
A. 0
B. 2
C. 3
D. 4
E. 6
For least value,
4y^2 needs to be zero
we know,
x+y=2
take y=0; x will be 2
putting values in equation,
2^2 + 0 + 0
= 4
option D
07 Jun 2019, 04:15
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nick1816
The least value of \(x^2 + 5xy + 4y^2\) given x + y = 2 and x, y are non negative, is:
A. 0
B. 2
C. 3
D. 4
E. 6
A. 0
B. 2
C. 3
D. 4
E. 6
\(x^2 + 5xy + 4y^2 = (x+2y)^2 + xy\)
i.e. x+2y must be minimized for the minimum value of (x+2y)^2 which will be minimum if y is minimum and \(y_{min} = 0\)
i.e. x = 2 for \(y_{min} = 0\) because x+y = 2
so \(x^2 + 5xy + 4y^2 = (x+2y)^2 + xy = (2+0)^2 +2*0 = 4\)
Answer: Option D
