S ( t ) = 2 ⋅ t ⋅ ( 7 − t 2 ) = a r e a o f r e c t a n g l e S ’ ( t ) = 2 ⋅ ( 7 − t 2 ) − 4 ⋅ t 2 = 14 − 6 ⋅ t 2 i s t h e d e r i v a t i v e S ’ ( t ) = 14 − 6 ⋅ t 2 = 0 w e f i n d z e r o s o f t h e d e r i v a t i v e t 0 = 7 3 − u n i q u e p o s i t i v e r o o t S ′ ′ ( t 0 ) = − 12 ⋅ t 0 = − 12 7 3 < 0 , t h e s e c o n d d e r i v a t i v e i s n e g a t i v e , t h e r e f o r e t 0 i s m a x i m a l p o i n t B a s e o f t h e l a r g e s t r e c t a n g l e i s a 0 = 2 ⋅ 7 3 T h e h e i g h t i s h 0 = 7 − 7 3 = 14 3 S(t)=2\cdot t\cdot \ (7-t^2) =\space area\space of\space rectangle\\
S’(t)=2\cdot (7-t^2) -4\cdot t^2=14-6\cdot t^2\space is\space the \space derivative\\
S’(t)=14-6\cdot t^2=0 \space \\ we \space find\space zeros\space of\space the \space derivative\\
t0=\sqrt\frac{7}{3} -\space unique \space positive \space root\\
S''(t0)=-12\cdot t0=-12\sqrt\frac{7}{3}<0, \space the \space second \space derivative\\
is\space negative,\space therefore\space t0\space is \space maximal\space point\\
Base\space of\space the \space largest\space rectangle \space is\space
a0=2\cdot\sqrt\frac{7}{3}\\
The\space height\space is \space h0=7-\frac{7}{3}=\frac{14}{3} S ( t ) = 2 ⋅ t ⋅ ( 7 − t 2 ) = a re a o f rec t an g l e S ’ ( t ) = 2 ⋅ ( 7 − t 2 ) − 4 ⋅ t 2 = 14 − 6 ⋅ t 2 i s t h e d er i v a t i v e S ’ ( t ) = 14 − 6 ⋅ t 2 = 0 w e f in d zeros o f t h e d er i v a t i v e t 0 = 3 7 − u ni q u e p os i t i v e roo t S ′′ ( t 0 ) = − 12 ⋅ t 0 = − 12 3 7 < 0 , t h e seco n d d er i v a t i v e i s n e g a t i v e , t h ere f ore t 0 i s ma x ima l p o in t B a se o f t h e l a r g es t rec t an g l e i s a 0 = 2 ⋅ 3 7 T h e h e i g h t i s h 0 = 7 − 3 7 = 3 14
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