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embizze | High School Teacher | (Level 2) Educator Emeritus

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We are given a rectangle with perimeter 50 inches and we are asked to find the dimensions so that the area is maximum.

Let the dimensions be l x w (length by width.) Since opposite sides of a rectangle are congruent, and the perimeter is the sum of the sides we have:

50= 2l + 2w

We solve this for one of the variables:

2w=50-2l ==> w=25-l

Now the area is the product of the length and the width or A=lw. We have an expression for w in terms of l so we have:

A=l(25-l) or `A=25l-l^2 `

This is a quadratic in l -- the graph is a parabola opening down so the maximum occurs at the vertex.

There are several ways to find the vertex -- it is midway between the x-intercepts (if any); the x-coordinate is `x=(-b)/(2a) ` if the quadratic is written in standard form `ax^2+bx+c=0 ` , etc...

Here the vertex is at (12.5,12.5) so l=w=12.5


The dimensions are 12.5 x 12.5 inches for an area of 156.25 square inches.


Since this was a multiple choice question, you could also check that the dimensions given yield a rectangle with perimeter 50, and of those select the one with the most area.

The graph of the area function:

If you have calculus, take the first derivative of the area function -- where the derivative is zero is a critical point and will be the maximum.


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