Question

Maximize Q=xy^2, where x and y are positive numbers, such that x+y^2=4

Answer 1

`L(x,y,\lambda)=xy^2+\lambda(x+y^2-4)`

`\begin{cases}L_x=y^2+\lambda=0&(1)\\L_y=2xy+2\lambda y=0&(2)\\L_\lambda=x+y^2-4=0&(3)\end{cases}`

Multiply (1) by
`2y`, then subtract (2) from that result:


`2y^3+2\lambda y=0`

`\implies(2y^3+2\lambda y)-(2xy+2\lambda y)=0`

`y^3-xy=0`

`y(y-\sqrt x)(y+\sqrt x)=0`

which means
`y=0`,
`y=-\sqrt x`, or
`y=\sqrt x`. But
`x>0`, so we only take
`y=\sqrt x`.

Substitute into (3) and solve for
`x`.


`x+(\sqrt x)^2=4\implies2x=4\implies x=2\implies y=\sqrt2`

So a
`Q` attains a maximum at
`(2,\sqrt2)`, giving a maximum value of
`2(\sqrt2)^2=4`.