Exams Joint Entrance Examination Mathematics Application of Derivatives Question 59 Question 59 of 212 Medium NUMERICAL
Let A A A be the region enclosed by the parabola y 2 = 2 x {y}^{2}=2x y 2 = 2 x and the line x = 24 x=24 x = 24 . Then the maximum area of the rectangle inscribed in the region A A A is \_\_\_\_\_\_\_\_.
Answer: 128
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y 2 = 2 x a = ( 1 2 ) \begin{gathered} {y}^{2}=2x \\ a=(\frac{1}{2}) \end{gathered} y 2 = 2 x a = ( 2 1 )
A ( t ) = 2 t × ( 24 − t 2 2 ) A = 48 t − t 3 \begin{gathered} A(t)=2t\times (24-\frac{{t}^{2}}{2}) \\ A=48t-{t}^{3} \end{gathered} A ( t ) = 2 t × ( 24 − 2 t 2 ) A = 48 t − t 3
d A d t = 48 − 3 t 2 48 − 3 t 2 = 0 3 t 2 = 48 t 2 = 16 t = ± 4 \begin{gathered} \frac{dA}{dt}=48-3{t}^{2} \\ 48-3{t}^{2}=0 \\ 3{t}^{2}=48 \\ {t}^{2}=16 \\ t=\pm 4 \end{gathered} d t d A = 48 − 3 t 2 48 − 3 t 2 = 0 3 t 2 = 48 t 2 = 16 t = ± 4
A ( 4 ) = 48 × 4 − 4 3 = 192 − 64 A ( 4 ) = 128 \begin{gathered} A(4)=48\times 4-{4}^{3} \\ =192-64 \\ A(4)=128 \end{gathered} A ( 4 ) = 48 × 4 − 4 3 = 192 − 64 A ( 4 ) = 128 Answer the question to unlock