# Position of vertices of right triangle inscribed on $x^2+4y^2=1$ with maximum area using Lagrange Multipliers

I am asked to find, using Lagrange multipliers, the position of the vertices of a right triangle inscribed on $x^2+4y^2=1$ that has the maximum area. The two legs of the triangle (which are not the hypothenuse) are parallel to the elypsis' axis.

What I have so far is:

$$2y = 2 \lambda x\\ 2x = 8 \lambda y\\ x^2+4y^2=1$$

By solving these equations it is known that both $\lambda$ and $x$ cannot be zero. I got minimum points to be:

$$f \left( \mp \frac{1}{\sqrt{2}} , \pm \frac{1}{2\sqrt{2}} \right) = \frac{1}{2}$$

Since the textbook has no answer, I want to make sure if that is correct.

EDIT

Let's say that we need to find a rectangle given the same conditions. Are the critical points the same (which would lead to an area of $1$)?

EDIT 2

Cheers.

• Is one vertex the origin and the two (short) legs parallel to the $x$ and $y$ axes? – David G. Stork May 26 '16 at 23:58
• @DavidG.Stork I just edited the question David: the two catheti are parallel to the axis of the elypsis. Thank you. – bru1987 May 27 '16 at 0:00
• Whew... that makes the problem much simpler. – David G. Stork May 27 '16 at 0:03
• Hahahahh yes, sorry about that! – bru1987 May 27 '16 at 0:04

You wish to maximize $A = f(x,y) = {1 \over 2} (2 x) (2 y) = 2 x y$ subject to the constraint $g(x,y) = x^2 + 4 y^2 - 1 = 0$.

Form the Lagrangian:

${\cal L}(x,y,\lambda) = f(x, y) - \lambda g(x,y) = 2 x y - \lambda (x^2 + 4 y^2 - 1)$.

Take the three derivatives and set them to zero:

• ${\partial {\cal L}(x,y,\lambda) \over \partial x} = 2 y - 2 \lambda x = 0$
• ${\partial {\cal L}(x, y, \lambda) \over \partial y} = 2 x - \lambda 8 y = 0$
• ${\partial {\cal L}(x, y, \lambda) \over \partial \lambda} = -(x^2 + 4 y^2 - 1) = 0$

Then solve these three simultaneous equations for $x,y, \lambda$, then plug in $x$ and $y$ to the formula for area.

I get $x = -{1 \over \sqrt{2}}, y = -{1 \over 2 \sqrt{2}}$ and thus $A = {1 \over 2}$.

If the problem is instead to maximize the area of a rectangle, then the area function is doubled: $f(x,y) = (2 x)(2 y)$ and the optimum turns out to be the same geometry (and of course twice the area), even though the solution $\lambda$ differs from the triangle case.

• would you mind double checking your partial derivatives? I seem to have found $2 \times$ what you've got for all of them... – bru1987 May 27 '16 at 0:22
• Not really David. Let me check everything again... – bru1987 May 27 '16 at 0:24
• David would you mind checking the image I uploaded? It seems that it is actually $1/2$, agreed? – bru1987 May 27 '16 at 0:35
• I didn't answer "yes" to that as you may check on the discussion above. But nevermind, it was just a misunderstanding. Thank you anyways. – bru1987 May 27 '16 at 0:40