Question 1014615
<font face="Times New Roman" size="+2">


Let *[tex \Large w] represent the width and *[tex \Large l] represent the length.  Since *[tex \Large P\ =\ 2l\ +\ 2w]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ +\ w\ =\ \frac{P}{2}],


In other words:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ \frac{P}{2}\ -\ w]


If the length is *[tex \Large a] units more than *[tex \Large b] times the width, then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ a\ +\ bw]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{P}{2}\ -\ w\ =\ a\ +\ bw]


Substitute your given values for *[tex \Large a], *[tex \Large b], and *[tex \Large P] and then solve for *[tex \Large w].  Finally, calculate *[tex \Large a\ +\ bw]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

</font>