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26 May 2017, 04:43
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If \(f(x,y) = \frac{10x}{x+2y} + \frac{20y}{2x+y}\), where \(0 < x < y\), which of the following could be the value of \(f(x, y)\)?
A. 9
B. 10
C. 19
D. 20
E. 23
A. 9
B. 10
C. 19
D. 20
E. 23
26 May 2017, 04:46
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Bookmarks
Official Solution:
If \(f(x,y) = \frac{10x}{x+2y} + \frac{20y}{2x+y}\), where \(0 < x < y\), which of the following could be the value of \(f(x, y)\)?
A. 9
B. 10
C. 19
D. 20
E. 23
From \(f(x, y) = \frac{10x}{x+2y} + \frac{20y}{2x+y}\), there are 2 variables \((x, y)\), so it is difficult to find the value of \(f(x, y)\). If so, you have to use \(0 < x < y\),
(1) if \(x=0, f(0, y) = \frac{10(0)}{0+2y}+\frac{20y}{2(0)+y} = \frac{20y}{y} = 20\).
(2) if \(x=y, f(x, x) = \frac{10x}{x+2x}+\frac{20x}{2x+x} = \frac{10x}{3x}+\frac{20x}{3x} = \frac{30x}{3x} = 10\).
However, from \(0 < x < y\), \(x\) is the range between 0 and \(y\), excluding 0 and y, so \(f(x, y)\) also becomes the range between 10 and 20, excluding 10 and 20. In other words, the only option that satisfies \(10 < f(x, y) < 20\) is C, 19. The answer is C.
Answer: C
If \(f(x,y) = \frac{10x}{x+2y} + \frac{20y}{2x+y}\), where \(0 < x < y\), which of the following could be the value of \(f(x, y)\)?
A. 9
B. 10
C. 19
D. 20
E. 23
From \(f(x, y) = \frac{10x}{x+2y} + \frac{20y}{2x+y}\), there are 2 variables \((x, y)\), so it is difficult to find the value of \(f(x, y)\). If so, you have to use \(0 < x < y\),
(1) if \(x=0, f(0, y) = \frac{10(0)}{0+2y}+\frac{20y}{2(0)+y} = \frac{20y}{y} = 20\).
(2) if \(x=y, f(x, x) = \frac{10x}{x+2x}+\frac{20x}{2x+x} = \frac{10x}{3x}+\frac{20x}{3x} = \frac{30x}{3x} = 10\).
However, from \(0 < x < y\), \(x\) is the range between 0 and \(y\), excluding 0 and y, so \(f(x, y)\) also becomes the range between 10 and 20, excluding 10 and 20. In other words, the only option that satisfies \(10 < f(x, y) < 20\) is C, 19. The answer is C.
Answer: C
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