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

Find the maximum and minimum values of function y=2x3−15x2+36x+11

A
42;41
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
39;38
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
35;34
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
52;51
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 39;38
y=2x315x2+36x+11
dydx=6x230x+36
dydx=0=6x230x+36=0
x25x+6=0
x22x3x+6=0
(x3)(x2)=0
Roots of the quadratic equation are
x=2, 3

Now double differentiation of given equation
d2ydx2=12x30
At x=2, d2ydx2=6
Hence at x=2 there is maxima of value ymax=39
At x=3, d2ydx2=+6
Hence at x=3, there is minima of value ymin=38

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