Proving the Local Maxima of a Rectangle Partition with Two Parallel Lines

[clairaut]

A rectangle with length L and width W is cut into four smaller rectangles by two lines parallel to the sides. Find the maximum and minimum values of the sum of the squares of the areas of the smaller rectangles.

Unless I did incorrectly, the algebra is very very long...

HELP

 

[clairaut]

[A(x,y)]= 4x^2y^2 + 2L^2y^2 - 4Lxy^2 - 4wx^2y + 2w^2x^2 + 4wLxy - 2w^2Lx - 2wL^2y + w^2L^2

Constraint C= WL = area1 = constant

 

[pasmith]

Firstly, this belongs in the homework forum.
Secondly, this problem does not involve lagrange multipliers.
Thirdly, it's best to factorize A(x,y) thusly: <br /> A(x,y) = x^2 y^2 + x^2(W-y)^2 + (L - x)^2y^2 + (L-x)^2(W-y)^2 = (x^2 + (L-x)^2)(y^2 + (W-y)^2). Both factors are non-negative in the domain we're interested in, so the maximum of the product is the product of the maxima, and the minimum of the product is the product of the minima.

 

[clairaut]

Firstly, this belongs in the homework forum.
Secondly, this problem does not involve lagrange multipliers.
Thirdly, it's best to factorize A(x,y) thusly: <br /> A(x,y) = x^2 y^2 + x^2(W-y)^2 + (L - x)^2y^2 + (L-x)^2(W-y)^2 = (x^2 + (L-x)^2)(y^2 + (W-y)^2). Both factors are non-negative in the domain we're interested in, so the maximum of the product is the product of the maxima, and the minimum of the product is the product of the minima.

Thanks!

When i do the first and second derivative test, I can find the local minimums.

However, I can only deduce the local maximum without a formal derivative test. Is there a way to mathematically prove the local maxima?