Max?

Algebra Level 2

What is the value of x Y 2 − x Y + x xY^{2} - xY + x if the value of y is max in Y x 2 − Y x + Y = 1 Yx^2 - Yx + Y = 1 and x is a real number?

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Note: Round off your answer to the nearest hundredths.


The answer is 0.72.

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Christian Daang by Christian Daang , 20, Philippines

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1 solution

Christian Daang
Jan 26, 2015

S o l u t i o n : Solution:

By Solving the equation,

Y x 2 − Y x + Y = 1 Yx^{2} - Yx + Y = 1

Y x 2 − Y x = 1 − Y Yx^{2} - Yx = 1-Y

x 2 − x + 1 / 4 = 1 − Y Y + 1 4 x^{2} - x + 1/4 = \cfrac{1-Y}{Y} + \cfrac{1}{4}

( x − 1 / 2 ) 2 = 4 − 3 Y 4 Y (x-1/2)^{2} = \cfrac{4-3Y}{4Y}

x = Y + / − Y ( 4 − 3 Y ) 2 Y x = \cfrac{Y+/- \sqrt{Y(4-3Y)}}{2Y}

Since x is a real no., y should be in range: 0 < / = Y < / = 4 / 3 0 </= Y </= 4/3 So, m a x ( Y ) = 4 / 3 max(Y) = 4/3

∴ x Y 2 − x Y + x = ( 1 / 2 ) ( 4 / 3 ) 2 − ( 1 / 2 ) ( 4 / 3 ) + ( 1 / 2 ) = 8 / 9 − 2 / 3 + 1 / 2 = 13 / 18 \therefore xY^{2} - xY + x = (1/2)(4/3)^{2} - (1/2)(4/3) + (1/2) = 8/9 - 2/3 + 1/2 = 13/18 which is approximately equal to 0.72 \boxed{0.72}

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