Consider an isosceles triangle of fixed perimeter P.
a) If x equals the length of one of the two equal sides, express the area A as a function of x.
b) What is the domain of A?
1)$\displaystyle
A(x) = \frac{(P-2x)\sqrt{4Px-P^2}}{4}
$
(Note that A(x) above IS a function of x (and not x and P), since P is a fixed constant.)
2)$\displaystyle
\text{Domain of } A: \frac{P}{4} < x < \frac{P}{2}
$
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Hello, dartanian24!
Consider an isosceles triangle of fixed perimeter $\displaystyle P$.
a) If $\displaystyle x$ equals the length of one of the two equal sides,
. . express the area $\displaystyle A$ as a function of $\displaystyle x\quad{\color{red}\hdots\;\text{ and }P}$Let $\displaystyle b$ = base of the triangle.Code:A * /|\ / | \ / | \ x / | \ x / |h \ / | \ / | \ B *-------*-------* C ½b D ½b : - - - b - - - :
Let $\displaystyle h$ = height of the triangle.
Then: .$\displaystyle b + 2x \:=\:P \quad\Rightarrow\quad b \:=\:P-2x \;\;{\color{blue}[1]}\quad\Rightarrow\quad \frac{b}{2} \:=\:\frac{P - 2x}{2}\;\;{\color{blue}[2]}$
In right triangle $\displaystyle ADB\!:\;\;h^2 + \left(\frac{b}{2}\right)^2 \:=\:x^2
$
Substitute [2]: .$\displaystyle h^2 + \left(\frac{P-2x}{2}\right)^2 \:=\:x^2 \quad\Rightarrow\quad h^2 + \frac{P^2 - 4Px + 4x^2}{4} \:=\:x^2$
Multiply by 4: .$\displaystyle 4h^2 + P^2 - 4Px + 4x^2 \:=\:4x^2 \quad\Rightarrow\quad h^2 \:=\:\frac{4Px-P^2}{4}$
Hence: .$\displaystyle h \:=\:\frac{\sqrt{4Px-P^2}}{2}\;\;{\color{blue}[3]}$
We know that: .$\displaystyle A \:=\:\tfrac{1}{2}bh$
Subnstitute [1] and [3]: .$\displaystyle A \;=\;\tfrac{1}{2}(P-2x)\frac{\sqrt{4Px-P^2}}{2} $
Therefore: .$\displaystyle A \;=\;\frac{1}{4}(P-2x)\sqrt{4Px-P^2}$
You probably mean "range".b) What is the domain of $\displaystyle A$?
The area has a minimum of 0 . . . when $\displaystyle b = 0.$
The area is a maximum when the triangle is equilateral.
. . Then: .$\displaystyle 3x \:=\:P \quad\Rightarrow\quad x \:=\:\frac{P}{3}$
. . The area is: .$\displaystyle \frac{\sqrt{3}}{4}\left(\frac{P}{3}\right)^2 \;=\;\frac{\sqrt{3}}{36}P^2$
The range is: .$\displaystyle \left[0,\:\frac{\sqrt{3}}{36}P^2\right]$