Extrema of a multivariable function

[adartsesirhc]

Homework Statement

Show that among all parallelograms with perimeter $$l$$, a square with sides of length $$l/4$$ has maximum area. Do this using the second partials test, and then using Lagrange multipliers.

Homework Equations

Area of a parallelogram: $$A = absin\phi$$, where a and b are the lengths of two adjacent sides and $$\phi$$ is the angle between them.

Second partials test: $$D=f_{xx}f_{yy} - f_{xy}^{2}$$

Method of Lagrange multipliers: $$\nabla f=\lambda\nabla g$$

The Attempt at a Solution

To do it using the second partials test, I would have to reduce A to a function of two variables. I know this has to do with expressing $$\phi$$ as a function of $$a$$ and $$b$$. After this, I'm stuck.

For the Lagrange multipliers method, I can use $$A(a,b)$$ as the function to be maximized and use $$g(a,b)= 2a + 2b - l$$ as the constraint equation. However, I'm still stuck on how to reduce $$A$$ to a function of two variables. Any help?

 

[Defennder]

Actually you don't really have to take $$\phi$$ into consideration. Can you show that one particular critical point occurs when a=b? So geometrically that reduces the possibilities to either a square or a rhombus. Since the area function you want to maximise is given by $$ab\sin \phi$$ and a=b, it's easy to tell that the expression is maxed when $$\phi = \frac{\pi}{2}$$.