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March 6, 2015

March 6, 2015

Posted by **l** on Thursday, August 31, 2006 at 10:18pm.

28 inches

Well, finally a calculus problem.

Ok, we know that the area for a rectangle is

A=l*w and the perimeter is P=2(l+w)

In this problem P= 28, so let's express one of the dimensions in terms of the other an substitute into the area formula. Thus 28=2(l+w). Let's solve for l in terms of w, thus 14=l+w, or l=14-w When we substitute this into the area formula we get

A=(14-w)*w. So A=14w-w^2

Now find dA/dw and evaluate the critical points.

dA/dw = 14-2w and dA/dw = 0 means 14-2w=0

So w = 7 is a critical point. I'll let you verify that this is the max. (the second deriv is -2, what does that mean?)

Thus the rectangle of maximum area has w=7. If you put this back into the formula for the perimeter, you'll find that l=7 too. This means that the rectangle of max. area is a square with a side=P/4.

Considering all rectangles with the same perimeter, the square encloses the greatest area.

Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.

This should give you your answer.

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