y=20x^2+2x+1200 (x>12).求y最小值 用基本不等式解
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y=20x^2+2x+1200 (x>12).求y最小值 用基本不等式解
y=20x^2+2x+1200 (x>12).求y最小值 用基本不等式解
y=20x^2+2x+1200 (x>12).求y最小值 用基本不等式解
y=20x^2+2x+1200
=20(x^2+x/10)+1200
=20(x+1/20)^2+1200 -1/20
x>12时,y无最小值
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