Question 583299
<font face="Times New Roman" size="+2">


You don't need to yell.  We can read.


Let *[tex \Large C] be the constant sum of the base and height of a triangle.  Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C\ =\ b\ +\ h]


Solving for *[tex \Large h]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ C\ -\ b]


Substitute for *[tex \Large h] in the formula for the area of a triangle to create an Area formula as a function of *[tex \Large b].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(b)\ =\ \frac{b(C\ -\ b)}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(b)\ =\ \frac{Cb\ -\ b^2}{2}]


Take the first derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dA}{db}\ =\ \frac{C\ -\ 2b}{2}]


Set the first derivative equal to zero and solve


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C\ -\ 2b\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ \frac{C}{2}]


is a local extreme.


Take the second derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2A}{db^2}\ =\ -1]


which is negative for all values of the variable, hence the extreme is a maximum.


Substitute back into the original relationship between *[tex \Large h] and *[tex \Large b]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ C\ -\ \frac{C}{b}\ =\ \frac{C}{b}]


Hence the area is maximum when the base is equal to the height.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>