The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12−x^{2} such that the rectangle lies inside the parabola, is

a. |
20 |
b. |
18 |

c. |
32 |
d. |
36 |

The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12−x^{2} such that the rectangle lies inside the parabola, is

a. |
20 |
b. |
18 |

c. |
32 |
d. |
36 |

1 Answer

127 votes

f(a) = 2a (12 − a)^{2}_{ }

f ’(a) = 2(12 − 3a^{2})

maximum at a = 2

maximum area = f(2) = 32

127 votes

127