Question

Maximize: Q=4xy^2 where x and y are positive numbers such that x+y^2=6. so.. x=? and y=?

Answer 1

What you could do is differentiate any of the folliwing equations
Q = 4(-y^2+6)*y^2 or Q = 4x*(-x+6)

Answer 2

keep in mind that x and y are greater than 0

Answer 3

@anncaseb which one equation seems easier ?

Answer 4

@anncaseb you're free to ask anything, Im just trying to help you out

Answer 5

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

Answer 6

if we say that dQ/dy = 0 that means the function has a maximum or a minimum value

Answer 7

if we were taking the second equation we should fin out the same answer, lets take a look:

Answer 8

Q = 4x(-x +6)

Simplifying:

Q= -4x^2 +6x

Differentiating:

dQ/dx = -8x + 6

now when dQ/dx = 0
0 = -8x +6

Answer 9

solve for x and you will find out the maximum value
bye

Answer 10

or we could do lagrange multipliers ...

Answer 11

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 ...

Answer 12

lagrane multipliers really nice :)