wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let y=2x321x2+36x20, then (x(0,8)). The point where Maxima occurs :

A
1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
23
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
14
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 1
The extremum of a function is a point where a maximum or minimum value of the function is found.

The given information is:

y=2x321x2+36x20 and x(0,8)

Differentiating the function w.r.t.x and equating it to 0 to find the extremum points.

y=6x242x+36=0

6x242x+36=0

x27x+6=0

(x1)(x6)=0

So the two extremum points are x=1 and x=6.

To find whether the function has the maximum or minimum value at the given extremum point we find the double derivative of the function. If its is greater than zero at the extremum point then its a minimum point else a maximum point. The double derivative test fails if it is equal to zero.

So find the double derivative by differentiating y we get,

y′′=12x42

y′′(1)=1242=30

y′′(1)<0. This means that maxima occurs at the point x =1 ......Answer

y′′(6)=7242=30

y′′(6)>0.. This means that minima occurs at the point x=6.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Extrema
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon