Lagrange Multipliers

**jelloish** · May 1st 2009, 10:19 AM

I'm supposed to find the maximum value of f(x,y)=xy if (x,y) is a point on the ellipse 9x^2+16y^2=144.

I have that y=lambda18x and x=lambda32y but I'm not sure how to go about solving for x and y.

Thanks

**redsoxfan325** · May 1st 2009, 10:44 AM

Originally Posted by jelloish

I'm supposed to find the maximum value of f(x,y)=xy if (x,y) is a point on the ellipse 9x^2+16y^2=144.

I have that y=lambda18x and x=lambda32y but I'm not sure how to go about solving for x and y.

Thanks

$\nabla(xy) = \langle y,x\rangle$ and $\nabla(9x^2+16y^2) = \langle 18x,32y\rangle$

So $\langle y,x\rangle = \lambda\langle 18x,32y\rangle$

So you want to solve the system:
$y=18\lambda x$
$x=32\lambda y$
$9x^2+16y^2=144$

The best thing to do is to get rid of $\lambda$ first. So divide the first two equations to get: $\frac{x}{y} = \frac{16y}{9x} \implies 9x^2=16y^2$

Substituting into the third equation gives us $9x^2+9x^2=144 \implies x=2\sqrt{2} \implies y=\sqrt{\frac{9}{2}}=\frac{3\sqrt{2}}{2}$

So the maximum of $xy$ occurs when $x=2\sqrt{2}$ and $y=\frac{3\sqrt{2}}{2}$ , so the maximum value is $2\sqrt{2}\cdot\frac{3\sqrt{2}}{2} = \boxed{6}$

**Spec** · May 1st 2009, 12:11 PM

Another way to solve it is to switch to polar coordinates: $3x=12cos(t), 4y=12sin(t)$

Then you just need to find the maximum of a function of one variable:
$f(x,y)=xy=f(4cos(t),3sin(t))=12cos(t) sin(t)$

**redsoxfan325** · May 1st 2009, 12:49 PM

Originally Posted by Spec

Another way to solve it is to switch to polar coordinates: $3x=12cos(t), 4y=12sin(t)$

Then you just need to find the maximum of a function of one variable:
$f(x,y)=xy=f(4cos(t),3sin(t))=12cos(t) sin(t)$

True. The derivative of $12\cos(t)\sin(t)$ is $12\cos(2t)$ . Setting $12\cos(2t)=0$ gives you $t=\frac{\pi}{4}$ .

$12\cos(\pi/4)\sin(\pi/4) = 12\cdot\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2}}{2} = \boxed{6}$

which is the same answer as above.

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