Math 12 - the area of a rectangle The new equation created...

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Darren Pinder Math 12 Project #3 The problem tells us that the total perimeter of the rectangle is two. Therefore we find the equation of the perimeter of a rectangle, which is twice the width, plus twice the length is equal to the perimeter. We make this equation equal to two. Since we have to find the area in terms of width, we solve for length “l” in the equation giving us the length equal to one minus the width. We take this equation and plug it into the area equation that is width multiplied by the length is equal to
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Unformatted text preview: the area of a rectangle. The new equation created is area of a perimeter equal to the width subtracted by the width squared. To maximize the area we find the derivative of the equation which is dA/dw equal to one subtracted by two multiplied by the width. We make this equation equal to zero and solve for “w”. We get width equal to a half. This means that a half squared is equal to the maximum area, which is equal to one-fourth....
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