Homework Statement

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





Homework Equations

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



Second partials test: [tex]D=f_{xx}f_{yy} - f_{xy}^{2}[/tex]



Method of Lagrange multipliers: [tex]

abla f=\lambda

abla g[/tex]



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 [tex]\phi[/tex] as a function of [tex]a[/tex] and [tex]b[/tex]. After this, I'm stuck.



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