This is a word problem, that I need solved. I need t solved in the lowest terms.

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Among all rectangles that have a perimeter of 60, find the dimensions of the one whose area is largest.

Here's an algebraic solution.

x = width of rectangle

y = length of rectangle

x + y = 30 ==> y = 30 – x

Area, A = xy = x(30–x)

A(x) is a quadratic that graphs as a parabola opening down (-x^2);

so Vertex will be maximum point on parabola.

Zeros of A(x): x = 0 and x = 30

Axis of Symmetry: x = 30/2 = 15.

Vertex: (15,A(15))

So largest rectangle with perimeter of 60 has:

width = 15, and

length = 30 – 15 = 15.

It's a square.

If the perimeter is 60 then the special rectangle (a square) has sides of 15 each.

<< Yes, a square is a rectangle - it is a special case (or kind) of rectangle.>>

To get the area we multiply the length and width - i.e. 15 x 15 or 15 squared.

If you know about the square function, you understand squared number get larger very quickly.

Much more quickly than just multiplying two numbers.

So test it - what are the sides of a rectangle that isn't a square?

What about the rectangle that has two sides 14 - what are the other two sides?

Multiply that out and see if it is bigger or smaller than 15 squared.

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