標題:
Challenging optimization
發問:
A cuboid has total surface area of A. Find mathematically (not only state) the dimension of the cuboid so that its volume is a maximum and the corresponding volume. 更新: myisland8132, your calculation does not end. Although you are calculated that x = y = z, it does not immediately mean that x = y = z is a maximum. It may be maximum, minimum or stationary. You should do some checking to show that x = y = z is a maximum.
正如maximal_ideal_space所講，可以用AM>=GM黎做如下： Let the length, width, height be x, y, z respectively. First we know that A = 2[xy+yz+zx] and V = xyz we want to maximise V when A is fixed. Thus applying the AM>=GM, we get A / 2 = xy+yz+zx >= 3 [xy*yz*zx]^{1/3} = 3 [xyz]^{2/3} = 3 V^{2/3} Thus we have V <= (A / 6)^{3/2} Equality holds iff xy=yz=zx iff x=y=z.
其他解答:
正如maximal_ideal_space所講，可以用AM>=GM黎做如下： Let the length, width, height be x, y, z respectively. First we know that A = 2[xy+yz+zx] and V = xyz we want to maximise V when A is fixed. Thus applying the AM>=GM, we get A / 2 = xy+yz+zx >= 3 [xy*yz*zx]^{1/3} = 3 [xyz]^{2/3} = 3 V^{2/3} Thus we have V <= (A / 6)^{3/2} Equality holds iff xy=yz=zx iff x=y=z. Remark. When you are doing such problems with fixed sum or fixed product, and want to maximise the product or minimise the sum, then you may simply use AM>=GM to solve the optimization problems. 2009-01-31 17:32:10 補充： #003抄襲#002！！！ 請大家唔好投#003！！！|||||其實簡單的用AM-GM inequality也可以。|||||Let use Lagrange Multiplier Method to solve this problem Let length x, width y, height z Then total surface area = 2(xy+yz+zx)=A V=xyz So we want to maximize xyz subject to 2(xy+yz+zx)-A=0 Using Lagrange Multiplier Method Consider f(x,y,z)=xyz-λ[2(xy+yz+zx)-A] ?f/?x=yz-λ(2y+2z) ?f/?y=xz-λ(2x+2z) ?f/?z=xy-λ(2x+2y) ?f/?λ= 2(xy+yz+zx)-A Set them all equal to zero We have yz/(2y+2z)=xz/(2x+2z)=xy/(2x+2y) yz/(y+z)=xz/(x+z)=xy/(x+y) It can be eqaily shown that x=y=z So 2(xy+yz+zx)-A=0 =>6x^2=A x=SQRT(A/6) V=xyz=x^3=(A/6)^(3/2) 2009-01-16 18:34:30 補充： 正如maximal_ideal_space大大所知﹐這裡面有好高深的數理哲學呢。 正如周界相同﹐圓的面積最大。 http://hk.knowledge.yahoo.com/question/question?qid=7009011202123 您應該展示下你的實力嘛 2009-01-16 23:32:34 補充： 經貓朋提點﹐我知道答案了
Challenging optimization
發問:
A cuboid has total surface area of A. Find mathematically (not only state) the dimension of the cuboid so that its volume is a maximum and the corresponding volume. 更新: myisland8132, your calculation does not end. Although you are calculated that x = y = z, it does not immediately mean that x = y = z is a maximum. It may be maximum, minimum or stationary. You should do some checking to show that x = y = z is a maximum.
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最佳解答:正如maximal_ideal_space所講，可以用AM>=GM黎做如下： Let the length, width, height be x, y, z respectively. First we know that A = 2[xy+yz+zx] and V = xyz we want to maximise V when A is fixed. Thus applying the AM>=GM, we get A / 2 = xy+yz+zx >= 3 [xy*yz*zx]^{1/3} = 3 [xyz]^{2/3} = 3 V^{2/3} Thus we have V <= (A / 6)^{3/2} Equality holds iff xy=yz=zx iff x=y=z.
其他解答:
正如maximal_ideal_space所講，可以用AM>=GM黎做如下： Let the length, width, height be x, y, z respectively. First we know that A = 2[xy+yz+zx] and V = xyz we want to maximise V when A is fixed. Thus applying the AM>=GM, we get A / 2 = xy+yz+zx >= 3 [xy*yz*zx]^{1/3} = 3 [xyz]^{2/3} = 3 V^{2/3} Thus we have V <= (A / 6)^{3/2} Equality holds iff xy=yz=zx iff x=y=z. Remark. When you are doing such problems with fixed sum or fixed product, and want to maximise the product or minimise the sum, then you may simply use AM>=GM to solve the optimization problems. 2009-01-31 17:32:10 補充： #003抄襲#002！！！ 請大家唔好投#003！！！|||||其實簡單的用AM-GM inequality也可以。|||||Let use Lagrange Multiplier Method to solve this problem Let length x, width y, height z Then total surface area = 2(xy+yz+zx)=A V=xyz So we want to maximize xyz subject to 2(xy+yz+zx)-A=0 Using Lagrange Multiplier Method Consider f(x,y,z)=xyz-λ[2(xy+yz+zx)-A] ?f/?x=yz-λ(2y+2z) ?f/?y=xz-λ(2x+2z) ?f/?z=xy-λ(2x+2y) ?f/?λ= 2(xy+yz+zx)-A Set them all equal to zero We have yz/(2y+2z)=xz/(2x+2z)=xy/(2x+2y) yz/(y+z)=xz/(x+z)=xy/(x+y) It can be eqaily shown that x=y=z So 2(xy+yz+zx)-A=0 =>6x^2=A x=SQRT(A/6) V=xyz=x^3=(A/6)^(3/2) 2009-01-16 18:34:30 補充： 正如maximal_ideal_space大大所知﹐這裡面有好高深的數理哲學呢。 正如周界相同﹐圓的面積最大。 http://hk.knowledge.yahoo.com/question/question?qid=7009011202123 您應該展示下你的實力嘛 2009-01-16 23:32:34 補充： 經貓朋提點﹐我知道答案了
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