Question

y=(40x-x^2)/2怎么取他的最大值

Answer 1

y=(40x-x^2)/2

y=-1/2(x²-40)

y=-1/2[(x-20)²-400]

y=-1/2(x-20)²+200

∵x≤15

∴x-20≥5

∴(x-20)²≥25

∴-1/2(x-20)≤-25/2=-12.5

y≤-12.5+200=187.5

∴y=(40x-x^2)/2最小值为：187.5