Help with maths?
2016-10-08 9:13 am
A rectangle has a fixed area of 50 cm2. At what rate with respect to its length is its perimeter changing when length is 10 cm?
回答 (2)
2016-10-08 11:37 am
xy = 50
P = 2x+2y = 2x+100/x
dP/dx = 2-100/x^2
when x = 10 cm
dP/dx = 1 cm/cm
P = 2x+2y = 2x+100/x
dP/dx = 2-100/x^2
when x = 10 cm
dP/dx = 1 cm/cm
2016-10-08 9:27 am
A = lw
so, lw = 50
=> w = 50/l
Now, P = 2l + 2w
so, P = 2l + 2(50/l) => 2l + 100/l
or, P = 2l + 100l⁻¹
Hence, dP/dl = 2 - 100l⁻² => 2 - 100/l²
When l = 10 we have dP/dl = 2 - 100/100 => 1
:)>
so, lw = 50
=> w = 50/l
Now, P = 2l + 2w
so, P = 2l + 2(50/l) => 2l + 100/l
or, P = 2l + 100l⁻¹
Hence, dP/dl = 2 - 100l⁻² => 2 - 100/l²
When l = 10 we have dP/dl = 2 - 100/100 => 1
:)>
收錄日期: 2021-04-21 23:34:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161008011335AA7eCi4