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 :D

Printable View

- May 1st 2009, 10:19 AMjelloishLagrange Multipliers
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 :D - May 1st 2009, 10:44 AMredsoxfan325
$\displaystyle \nabla(xy) = \langle y,x\rangle$ and $\displaystyle \nabla(9x^2+16y^2) = \langle 18x,32y\rangle$

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

So you want to solve the system:

$\displaystyle y=18\lambda x$

$\displaystyle x=32\lambda y$

$\displaystyle 9x^2+16y^2=144$

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

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

So the maximum of $\displaystyle xy$ occurs when $\displaystyle x=2\sqrt{2}$ and $\displaystyle y=\frac{3\sqrt{2}}{2}$, so the maximum value is $\displaystyle 2\sqrt{2}\cdot\frac{3\sqrt{2}}{2} = \boxed{6}$ - May 1st 2009, 12:11 PMSpec
Another way to solve it is to switch to polar coordinates: $\displaystyle 3x=12cos(t), 4y=12sin(t)$

Then you just need to find the maximum of a function of one variable:

$\displaystyle f(x,y)=xy=f(4cos(t),3sin(t))=12cos(t) sin(t)$ - May 1st 2009, 12:49 PMredsoxfan325
True. The derivative of $\displaystyle 12\cos(t)\sin(t)$ is $\displaystyle 12\cos(2t)$. Setting $\displaystyle 12\cos(2t)=0$ gives you $\displaystyle t=\frac{\pi}{4}$.

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

which is the same answer as above.