Optimizing Area: Finding the Minimum Value of Sum of Squares for Cut Rectangles

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The discussion revolves around optimizing the sum of the squares of the areas of four smaller rectangles cut from a larger rectangle. The initial approach incorrectly assumes each smaller rectangle's area is simply one-fourth of the total area. A more accurate method involves defining the areas of the smaller rectangles based on the lengths of the cuts along the rectangle's sides. The conversation suggests using Lagrange multipliers for optimization, although the implementation details are unclear. Participants are seeking clarification and guidance on the correct mathematical approach to solve the problem.
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I am having a little bit of trouble with one of my math problems.

a) A rectanglewith length L and width W is cut into four smaller rectangles by two lines parallel to the sides. Find the minimum value of the sum of the squares of the areas of the smaller rectangles.

b) Show that the maximum of the sum of the squares of the areas occurs when cutting lines correspond to sides of the rectangle (so that there is only one rectangle).

i started part a) and this is what i got so far:

A=LW , A(small)=a , a=1/4*LW

so da/dL = (1/4)*W and da/dW= (1/4)*L

and the sum of the squares of all these are equal to 0 so
4[1/16(L^2+W^2)] = 0
1/4(L^2+W^2) = 0

I don't think that this is right though, can anyone help me out here?

thanks
 
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after thinking about it for a bit, i am now wondering if it would be easier if i were to try to use lagrange multipliers to solve this, but I am not too sure how i would implement them.
 
matpo30 said:
A=LW , A(small)=a , a=1/4*LW

Where're you getting this statement? Particularly that the area of the smaller rectangles is simply 1/4 the total area?

Let l be the length to the cut along the L side and w the length to the cut along the W side. Then,

a1 = lw
a2 = l(W-w)
a3 = (L - l)w
a4 = (L - l)(W - w)

and a1 + a2 + a3 + a4 = A (Note: this last equation won't help you).

See if you can't get that to work.

--Justin
 
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