anonymous
  • anonymous
Maximize: Q=4xy^2 where x and y are positive numbers such that x+y^2=6. so.. x=? and y=?
Calculus1
jamiebookeater
  • jamiebookeater
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rock_mit182
  • rock_mit182
What you could do is differentiate any of the folliwing equations Q = 4(-y^2+6)*y^2 or Q = 4x*(-x+6)
rock_mit182
  • rock_mit182
keep in mind that x and y are greater than 0
rock_mit182
  • rock_mit182
@anncaseb which one equation seems easier ?

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rock_mit182
  • rock_mit182
@anncaseb you're free to ask anything, Im just trying to help you out
rock_mit182
  • rock_mit182
in order to find the maximum value we need to simplify one of the equations. lets take this one: Q = 4(-y2+6)*y^2 simplifiying: Q = -4y^4+ 24y^2 now you have to differentiate: dQ/dy = -16y^3 + 48y
rock_mit182
  • rock_mit182
if we say that dQ/dy = 0 that means the function has a maximum or a minimum value
rock_mit182
  • rock_mit182
if we were taking the second equation we should fin out the same answer, lets take a look:
rock_mit182
  • rock_mit182
Q = 4x(-x +6) Simplifying: Q= -4x^2 +6x Differentiating: dQ/dx = -8x + 6 now when dQ/dx = 0 0 = -8x +6
rock_mit182
  • rock_mit182
solve for x and you will find out the maximum value bye
amistre64
  • amistre64
or we could do lagrange multipliers ...
amistre64
  • amistre64
Q(x,y) = 4xy^2 Lg(x,y) = L(x+y^2-6) Qx = 4y^2 Qy = 8xy Lgx = L(1) Lgy = L(2y) 4y^2 = L 2Ly = 8xy ; L = 4x so y^2 = x 2x-6 = 0, etc ...
rock_mit182
  • rock_mit182
lagrane multipliers really nice :)

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