Question

the maximum value of y=25x^2+5-10x

Answer 1

Answer:

For maximum and minimum value, we can put dxdy=0.

or dxdy=50x−10

Put, dxdy=0,∴x=51

Further, dx2d2y=50

or dx2d2y has positive value at x=51. 

Therefore, y has minimum value at  x=51.

Substituting x=51 in given equation, we get

ymin=25(51)2+5−10(51)=4