Let x , x x, x x , x and y y y will be the sides of a isosceles triangle. If P = x + x + y , P=x+x+y, P = x + x + y , then y = P − 2 x . y=P-2x. y = P − 2 x .
The area S S S of the triangle is
S = p ( p − x ) ( p − x ) ( p − y ) S=\sqrt{p(p-x)(p-x)(p-y)} S = p ( p − x ) ( p − x ) ( p − y )
S = P 2 ( P 2 − x ) ( P 2 − x ) ( P 2 − ( P − 2 x ) ) S=\sqrt{\dfrac{P}{2}(\dfrac{P}{2}-x)(\dfrac{P}{2}-x)(\dfrac{P}{2}-(P-2x))} S = 2 P ( 2 P − x ) ( 2 P − x ) ( 2 P − ( P − 2 x ))
S ( x ) = P 4 ( P − 2 x ) 4 x − P , P 4 < x < P 2 S(x)=\dfrac{\sqrt{P}}{4}(P-2x)\sqrt{4x-P}, \dfrac{P}{4}<x<\dfrac{P}{2} S ( x ) = 4 P ( P − 2 x ) 4 x − P , 4 P < x < 2 P Find the first derivative with respect to x x x
S ′ ( x ) = ( P 4 ( P − 2 x ) 4 x − P ) ′ S'(x)=(\dfrac{\sqrt{P}}{4}(P-2x)\sqrt{4x-P})' S ′ ( x ) = ( 4 P ( P − 2 x ) 4 x − P ) ′
= P 4 ( − 2 4 x − P + 4 ( P − 2 x ) 2 4 x − P ) =\dfrac{\sqrt{P}}{4}\big(-2\sqrt{4x-P}+\dfrac{4(P-2x)}{2\sqrt{4x-P}}\big) = 4 P ( − 2 4 x − P + 2 4 x − P 4 ( P − 2 x ) ) = P 2 4 x − P ( P − 4 x + P − 2 x ) =\dfrac{\sqrt{P}}{2\sqrt{4x-P}}(P-4x+P-2x) = 2 4 x − P P ( P − 4 x + P − 2 x )
= P ( P − 3 x ) 4 x − P =\dfrac{\sqrt{P}(P-3x)}{\sqrt{4x-P}} = 4 x − P P ( P − 3 x ) Find the critical number(s)
S ′ ( x ) = 0 = > P ( P − 3 x ) 4 x − P = 0 S'(x)=0=>\dfrac{\sqrt{P}(P-3x)}{\sqrt{4x-P}}=0 S ′ ( x ) = 0 => 4 x − P P ( P − 3 x ) = 0 Since P 4 < x < P 2 , \dfrac{P}{4}<x<\dfrac{P}{2}, 4 P < x < 2 P , we have the critical number
x = P 3 x=\dfrac{P}{3} x = 3 P If P 4 < x < P 3 , S ′ ( x ) < 0 , S ( x ) \dfrac{P}{4}<x<\dfrac{P}{3}, S'(x)<0,S(x) 4 P < x < 3 P , S ′ ( x ) < 0 , S ( x ) decreases.
If P 3 < x < P 2 , S ′ ( x ) > 0 , S ( x ) \dfrac{P}{3}<x<\dfrac{P}{2}, S'(x)>0,S(x) 3 P < x < 2 P , S ′ ( x ) > 0 , S ( x ) increases.
The function S ( x ) S(x) S ( x ) has a local maximum at x = P 3 . x=\dfrac{P}{3}. x = 3 P .
Since the function S ( x ) S(x) S ( x ) has the only extremum for P 3 < x < P 2 , \dfrac{P}{3}<x<\dfrac{P}{2}, 3 P < x < 2 P , then
the function S ( x ) S(x) S ( x ) has the absolute maximum for P 3 < x < P 2 \dfrac{P}{3}<x<\dfrac{P}{2} 3 P < x < 2 P at x = P 3 . x=\dfrac{P}{3}. x = 3 P .
x = P 3 , y = P 3 x=\dfrac{P}{3}, y=\dfrac{P}{3} x = 3 P , y = 3 P The equilateral triangle has the maximum area.
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