高数题在线求解,急
展开全部
(1)
P
=R-C
=10x-0.01x^2 -5x-200
=5x-0.01x^2-200
P' = 5-0.02x
P'=0
5-0.02x=0
x=250
P''=-0.02 <0 (max)
销量=250才能使总利润取得最大值
(2)
1)∫ x^5 dx =(1/6)x^6 +C
2) ∫8sinx dx = -8cosx+C
(3)
∫(0->1) (3+e^x) dx
=[3x+e^x]|(0->1)
=3+e-1
=2+e
(4)
∫(0->1) √x.(1+√x) dx
=∫(0->1) [√x+x] dx
= [(2/3)x^(3/2)+ (1/2)x^2]|(0->1)
=2/3+1/2
=7/6
P
=R-C
=10x-0.01x^2 -5x-200
=5x-0.01x^2-200
P' = 5-0.02x
P'=0
5-0.02x=0
x=250
P''=-0.02 <0 (max)
销量=250才能使总利润取得最大值
(2)
1)∫ x^5 dx =(1/6)x^6 +C
2) ∫8sinx dx = -8cosx+C
(3)
∫(0->1) (3+e^x) dx
=[3x+e^x]|(0->1)
=3+e-1
=2+e
(4)
∫(0->1) √x.(1+√x) dx
=∫(0->1) [√x+x] dx
= [(2/3)x^(3/2)+ (1/2)x^2]|(0->1)
=2/3+1/2
=7/6
展开全部
1. 总利润 L = R-C = 5x-0.01x^2-200 = -0.01(x^2-500x)-200
= 625-0.01(x-250)^2-200 = 425-0.01(x-250)^2,
x = 250 s时,利润最大为 425 元。
2. ∫x^5dx = (1/6)x^6 + C, ∫8sinxdx = -8cosx + C
3. ∫<0, 1> (3+e^x)dx = [3x+e^x]<0, 1> = 3+e^3 -1 = 2+e^3
4. ∫<0, 1>√x(1+√x)dx = ∫<0, 1>(√x+x)dx
= [(2/3)x^(3/2) + x^2/2]<0, 1> = 2/3 + 1/2 = 7/6
= 625-0.01(x-250)^2-200 = 425-0.01(x-250)^2,
x = 250 s时,利润最大为 425 元。
2. ∫x^5dx = (1/6)x^6 + C, ∫8sinxdx = -8cosx + C
3. ∫<0, 1> (3+e^x)dx = [3x+e^x]<0, 1> = 3+e^3 -1 = 2+e^3
4. ∫<0, 1>√x(1+√x)dx = ∫<0, 1>(√x+x)dx
= [(2/3)x^(3/2) + x^2/2]<0, 1> = 2/3 + 1/2 = 7/6
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
去百度搜题帮,拍照搜题
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询