1.

ate maximum or minimum value of the function ` y= 25x ^(2) +5- 10x `A. `y _(min)=4`B. ` y _(max) =8`C. `y _(min) =8`D. `y _(max) =4`

Answer» Correct Answer - A
`(dy)/(dx) =50x-10 =0`
`x-1/5implies (d^(2)y)/( dx ^(2)) =50gt 0`
So y is minimum at `x=1//5`
`y_(""(min))=25xx(1)/(25)+ 5-10 xx1/5 =4`


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