微積分問題

Suppose the two sides of the rectangular are a and 2*b where 2*b is also the diameter of the semi-circle.

We want to maximize 2*a*b+pi*b^2/2 subject to 2*a+(2+pi)*b=12.

Using Lagrange multiplier method,

Consider f(a,b) = 2*a*b+pi*b^2/2 +lambda*(12-(2*a+(2+pi)*b)),

Differentiate f(a,b) with respect to a and b, respectively, we get

2b-2lambda=0
and
2a+pi*b-(2+pi)*lambda=0

solve the two equations
a=b=lamda

then in equation

2*a+(2+pi)*b=12

we replace b=a and solve it to get a=b=12/(4+pi).

so the maximum area is

2*a*b+pi*b^2/2=72 which is obtained by using a=b=12/(4+pi) for the side lengths.

首先, 要畫個圖, 如果個圖錯左, 就成題錯了, 你對下你個圖跟我的圖一唔一樣,

我唔識在此DRAW, so 我的圖是

上面有個半圓, 連接住下面是一個矩形(長方形), 半圓的直徑是矩形的長,

let r = 半圓的半徑
d = 矩形的長
pi = p = 3.1416
A = 面積 = area
因周界= 12米 所以, 半圓的周界加上長方形的周界 就會等於十二
切記, 半圓的直徑不包括在內, 因周界的定義是周邊的長度, 內在的不包括

therefore 12 = (2rp)/2 + 2r +2d = 半圓周界加上矩形的底邊 (即 2r) 加上2條寛
12= rp +2r+2d
2d = 12-rp-2r

d = (12-rp-2r)/2

AREA = 半圓的面積 + 矩形面積
----> A = (pr^2)/2 +2rd ^2 =平方
therefore , sub d into the equation

A = (pr^2)/2 + 2r ( 12-rp-2r)/2
化簡 即加號後 的兩個 2 可約去

A = (pr^2)/2 + 12r-r^2p -2r^2
A = 12r-2r^2 - (pr^2)/2

then differentiate the above equation
A' = 12- 4r -pr
to maximize the A (area), the differentiate equation (A') must equal 0 (zero)

therefore A' =12 -4r -pr =0
12- 4r- pr=0
4r + pr =12
r ( 4+p) =12
therefore r = 12/(4+p)
----> i.e r = 1.680297461 metre
sub r back into the d equation to find d

d = (12-rp-2r)/2
----> therefore d =1.680297461
so, we can find out A , which A = (pr^2)/2 + 2rd
A = 10.08178476 metre square

你需要的答案是 , r, d , 和 A
只要清楚寫明 這三數, 便可得分.
希望你明白, 唔明再問
~完~