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}.