What is the maximum value of $f(x,y) = xy$ subjected to the constraint $4x^2 + 9y^2 = 32?$

Question

I know the process and all but am not sure what to plug my final $y$ value into.

What I did:

$$∇f(x,y)=λ∇g(x,y),$$ where $$g(x,y) = 4x^2 + 9y^2 = 32.$$

I ended up with $\lambda = \frac{1}{8x}$, $λ = \frac{1}{18y}$, plugged both into each other and received $$y = \frac{4x}{9}.$$

I then substituted $y=4x/9$ into the constraint function and solved for $x$, obtaining $$x=\pm \sqrt\frac{72}{13}.$$

Now I'm not sure if that value is correct, but whether it is or not, what do I plug it into to find the maximum? The original $f(x,y)$ function? If so, I still have an $x$ though so I wouldn't get a value?

Answer 1

Second Approach:

The objective function is $$f(x,y)=xy+ \lambda(4x^2+9y^2-32).$$ Therefore, $\frac{\partial}{\partial x}f(x,y) = 0 \implies y + 8\lambda x = 0$, and $\frac{\partial}{\partial y}f(x,y) = 0 \implies x + 18\lambda y = 0$. Consequently, $\lambda = \pm \frac{1}{12}$ and $y=\pm \frac{2}{3}x$. Plugging in the value of $y$ in the constraint results in $x = \pm 2 \implies y = \pm \frac{4}{3}$ and $xy = \pm 8/3$. You may verify that $xy = 8/3$ is the maximum through Hessian matrix.

Answer 2

Show that $$xy\le \frac{8}{9}$$ and the equal sign holds if $$x=-2,y=-\frac{4}{3}$$

Answer 3

You need to use the second order optimal conditions. For that you will need the hessian of $f$ and then check if for all the critical points found the hessian is sdp or sdn.

Check here: https://en.wikipedia.org/wiki/Second_partial_derivative_test

and if you want a more formal explanation: http://www.numerical.rl.ac.uk/people/nimg/course/lectures/raphael/lectures/lec10slides.pdf

Answer 4

First approach: Using mean inequality, $$\frac{4x^2+9y^2}{2} \ge\sqrt{36x^2 y^2} = 6|xy|\ge 6xy \implies xy \le \frac{4x^2+9y^2}{12}=\frac{8}{3}.$$

Answer 5

$x = a\cos t\\ y = b\sin t$

We find $a,b$ when we put the ellipse in standard from.

$a = \sqrt {\frac {32}{4}}\\ b = \sqrt {\frac {32}{9}}$

$xy = ab\cos t\sin t = \frac {ab}{2} \sin 2t$

Maximizing $xy \implies \sin 2t = 1$

$xy = \frac {ab}{2} = \frac {8}{3}$

What went wrong....

$f(x,y,\lambda) = xy - \lambda (4x^2 + 9y^2 - 32)\\ \frac {\partial f}{\partial x} = y - 8\lambda x = 0\\ \frac {\partial f}{\partial y} = x - 18\lambda y = 0\\ 4x^2 + 9y^2 = 32$

$x = 18 \lambda y$

substitute

$y = 144 \lambda^2 y\\ \lambda = \pm \frac {1}{12}$

Positive will maximize, negative will minimize.

$y = \frac {2}{3} x$

This appears to be your problem.

Answer 6

An attempt without using the Lagrange multipliers:

Ellipse: $4x^2 +9y^2 = 32;$

Family of hyperbolas : $c= xy.$

$4x^2 + 9(c^2/x^2) = 32.$

$4x^4 + 9c^2 - 32x^2=0$

$v:= x^2;$

$4v^2 -32v +9c^2= 0.$

Discriminant: $(32)^2 - (16)(9)c^2 \ge 0.$

$(4)(16) -9c^2 \ge 0.$

$c^2 \le (8/3)^2.$

In the1st and 3rd quadrant we have $c>0.$

$\rightarrow$: $\max (c) =8/3.$