Find all extrema for $f(x,y) = 3xy$ subject to the constraint $4x^2 + 2y = 48$

Question

Find all extrema for $f(x,y) = 3xy$ subject to the constraint $4x^2 + 2y =48$.

I put it into the form of:

$3xy - \lambda(4x^2 +2y -48) = F(x,y,\lambda)$

$3xy - 4x^2\lambda -2y\lambda + 48\lambda$

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$F_x = 3y - 8x\lambda = 0$

$F_y = 3x - 2\lambda = 0$

$F_\lambda = 4x^2 - 2y + 48 = 0$

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I'm just learning this so beg my pardon, but am I allowed to solve for $y$ in $F_x$ or must I solve for $y$ in $F_y$ (if so then I don't know how since there is no y in $F_y$).

EDIT:

Thanks to the comments, I understand now:

$y = (8x\lambda) / 3$ and $x = (2\lambda)/3$.

Then using substitution,

$4((2\lambda)/3)^2 - 2((8x\lambda)/3) + 48 = 0$.

and then solve for $\lambda$ and continue to solve for $x$ and $y$

$\blacksquare$

Answer 1

You seem to be getting the hang of Lagrange multipliers, so here's another approach that works nicely because we're lucky with the constraint.

Saying that $4x^2 + 2y = 48$ is equivalent to saying that $y = 24 - 2x^2$. This is where we're lucky; we can solve the equation of the constraint for one of the variables. This means that any point subject to the constraint are of the form $(x, 24 - 2x)$. The intuition here is that instead of following the curve $4x + 2y = 48$ and look for maxima, we follow the $x$-axis, and for each point we look over to the corresponding point on the curve and see what the function value is over there.

Now we have a function of one variable, namely $f(x, 24-2x^2) = 3x(24-2x^2) = 72x - 6x^3$, which is easily differentiated and found to have two extrema, at $x = \pm 2$. Since the third degree term has a negative sign, the $-2$-solution is a minimum, and the $+2$-solution is a maximum. The exact coordinates are obtained by using the constraint to figure out $y$ again:

1. Coordinates of minimum: $(-2, 24 - 2\cdot (-2)^2) = (-2, 16)$. Value: $f(-2, 16)= -96$.
2. Coordinates of maximum: $(2, 24 - 2\cdot 2^2) = (2, 16)$. Value: $f(2, 16)= 96$.