A particle is moving in a straight line such that its distance S at any time t is given by S = t 4 4 - 2 t 3 + 4 t 2 - 7 . Find when its velocity is maximum and acceleration minimum.

S = t 4 4 - 2 t 3 + 4 t 2 - 7 Velocity, v = S ' = t 3 - 6 t 2 + 8 t Acceleration a = S ' ' = dv dt Therefore, a = S ' ' = 3 t 2 - 12 t + 8 To find minimum or maximum velocity, consider dv dt = 0 ⇒ 3 t 2 - 12 t + 8 = 0 Solving the above equation, we get, t = 2 ± 2 3 d 2 v dt 2 = 6 t - 12 6 ( 2 + 2 3 ) - 12 > 0 minima 6 ( 2 - 2 3 ) - 12 < 0 maxima Hence, at t = 2 - 2 3 velocity is maximum To find minimum or maximum velocity, consider d a dt = 0 So, 6 t - 12 = 0 ⇒ t = 2 > 0 at t = 2 acceleration is minimum.

E M B I B E

MATHEMATICS CLASS XII VOLUME-1>Chapter 18 - Maxima and Minima>EXERCISE>Q 39

HARD

12th CBSE

IMPORTANT

Earn 100

A box of constant volume $c$ is to be twice as long as it is wide. The material on the top and four sides cost three times as much per square meter as that in the bottom. What are the most economic dimensions?

Important Questions on Maxima and Minima

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 18 - Maxima and Minima>EXERCISE>Q 40

The sum of the surface areas of a sphere and a cube is given. Show that when the sum of their volumes is least, the diameter of the sphere is equal to the edge of the cube.

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 18 - Maxima and Minima>EXERCISE>Q 41

A given quantity of metal is to be cast into a half cylinder with a rectangular base and semicircular ends. Show that in order that the total surface area may be minimum the ratio of the length of the cylinder to the diameter of its semi-circular ends is

π : π + 2

.

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 18 - Maxima and Minima>EXERCISE>Q 42

The strength of a beam varies as the product of its breadth and square of its depth. Find the dimensions of the strongest beam which can be cut from a circular log of radius