作业帮若正数ab满足ab+2a+b=48则ab的最大值与2a+b的最小值分别是
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若正数ab满足ab+2a+b=48则ab的最大值与2a+b的最小值分别是
若正数ab满足ab+2a+b=48则ab的最大值与2a+b的最小值分别是
若正数ab满足ab+2a+b=48则ab的最大值与2a+b的最小值分别是
令ab = n
b = n/a
n + 2a + n/a =48
(1+1/a)n +2a =48
n = (48-2a)/(1+1/a) = 2(24a-a^2)/(a+1)=2(-a^2-2a-1 + 26a +26 -25) / (a+1)
= 2[26 - (a+1) - 25 / (a+1)] = 52 - 2 [(a+1) + 25/(a+1)...
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令ab = n
b = n/a
n + 2a + n/a =48
(1+1/a)n +2a =48
n = (48-2a)/(1+1/a) = 2(24a-a^2)/(a+1)=2(-a^2-2a-1 + 26a +26 -25) / (a+1)
= 2[26 - (a+1) - 25 / (a+1)] = 52 - 2 [(a+1) + 25/(a+1)]
在 (a+1) = 25/(a+1) 时
n有最大值，即 a+1 = 5, a =4时, n = 32
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令2a+b = m
b = m - 2a
ab = a(m-2a)
a(m-2a) + m =48
(a+1)m - 2a^2 =48
m = (2a^2+48)/(a+1) =2(a^2+24)/(a+1) = 2(a^2+2a+1 - 2a -2 + 25)/(a+1)
=2[-2 + (a+1) + 25/(a+1)] = -4 +2[(a+1) + 25/(a+1)]
在 (a+1) = 25/(a+1) 时
m有最小值，即 a+1 = 5, a =4时, m = 16
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ab + 2a +b = 48
(a+1)(b+2) = 50
(1+a)(1+b/2) = 25 = 5*5
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