SOLUTION: maximize Q=xy where x and y are positive numbers such that x+6y^2=2
The maximum value of Q is___ and occurs at x=__ and y=__
Algebra.Com
Question 1061119: maximize Q=xy where x and y are positive numbers such that x+6y^2=2
The maximum value of Q is___ and occurs at x=__ and y=__
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
x +6y^2 = 2
:
6y^2 = -x +2
:
y^2 = (-x/6) + (1/3)
:
y = ( (-x/6) + (1/3) )^(1/2)
:
this is a parabola that opens to the left
:
the y intercept is (1/3)^(1/2) and x intercept is 2
:
Here is the graph showing the positive values for x and y
:
:
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max value of Q is 2 * (1/3)^(1/2) = 1.1547 where x=2, y = (1/3)^(1/2)
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