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Find minimum value `y=25x^(2)-10x+5`. |
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Answer» For maximum/ minimum value `(dy)/(dx)=0 rArr50x-10rArrx=(1)/(5)` Now at `x=(1)/(5),(d^(2)y)/(dx^(2))=50`, which is positive So `y_(min)=25((2)/(5))^(2)-10((1)/(5))+5=1-2+5=4` |
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