Question 345735
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Set up the specific perimeter equation for this rectangle:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2l\ +\ 2w\ =\ 210]


Solve for *[tex \Large w]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ 105\ -\ l]


Substitute this expression for *[tex \Large w] into the area formula creating an expression for Area as a function of length:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(l)\ =\ l(105\ -\ l)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(l)\ =\ 105l\ -\ l^2]


Ordinarily a polynomial function would have the set of all real numbers as a domain, but this is a function that describes a physical object, and if *[tex \Large l\ \leq\ 0] or *[tex \Large l\ \geq\ 105], then a physical rectangle no longer exists because you cannot have a 0 or negative area for a real rectangle.


Hence the domain is the interval *[tex \Large \left(0,\,105\right)]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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