运筹学 习题一道关于线性规划的题.A company produces two types of cowboy hats

运筹学 习题
一道关于线性规划的题.
A company produces two types of cowboy hats.Atype 1 hat requires twice as much labour time as a type 2.If all the available labour time in producted to type 2 alone,the company can produce a total of 400 type 2 hats a day.the respective market limits for type1 and typ2 are 150 and 200 hats per day,respectively.the profit is $8 per type1 and $5 per type 2.determine the number of hats of each type that maximizes profit.
大致翻译:某公司生产两种帽子 type 1 花费的劳动时间是type 2 的两倍.如果单独生产type2的话一天能生产400个.但这两种帽子的市场限制是 type1 150个 type2 200个.他们的利润是 type1 ¥8 type2¥5.确定生产每个类型的帽子多少个能让利益最大化
同志们 给力啊.
求获利最大的产品生产计划
葛藤饭钟 1年前 已收到1个回答 举报

comicking 幼苗

共回答了22个问题采纳率:95.5% 举报

given fixed labor time per day. max total profit = max profit per labor time.
hat 1 has $4 profit/lt, hat 2 $5/lt. so max profit = max prodution of hat 2.
=> 200 hat 2, (400-200)/2=100 hat 1

1年前 追问

7

葛藤饭钟 举报

=。=线性规划题。。。 不是那种应用题 这么写我也会 我就是不会用课本上的那些式子~

举报 comicking

produce x hat1, y hat2 max 8x+5y s.t. x<=150, y<=200, x*2+y<=400 which method do u learn to solve LP problems? the easiest one is to take all corner points then plug in objective function to verify them one by one, but i'm not sure whether its the method that u are supposed to use.

葛藤饭钟 举报

其实=。=这个。。。。 我也都不知道我学的是哪种。。。我现在就是快小测了临时抱佛脚=。= 你就写你说的最简单的那种吧。。。。应该是那样的。。。坐标图我知道怎么弄。。。。来个具体过程就行了。。。。我最想要的是格式。。。。

举报 comicking

well the one i talked about is basically trial and error. take all constraints as equation, and solve for all corner points, then plug them in and compare the value of objective func. ur teacher wont like that though its easy to understand and efficient enough in simple problems. one generally used method to solve optimization (not restricted to linear problem) is to find out the stationary point of Lagrange function. http://en.wikipedia.org/wiki/Lagrange_multiplier let c1,c2,c3 be slack variables and l1, l2, l3 be the L-multiplier of the 3 constraints (usually lambda is used for mutiplier but hard to type here so i used l), respectively, the lagrange func L is L(x,y, c1, c2, c3, l1,l2,l3) = 8x+5y+l1(x+c1-150)+l2(y+c2-200)+l3(x*2+y+c3-400) solve the system: dL/dx = 0, dL/dy=0, dL/dc1=0, dL/dc2=0, dL/dc3=0, dL/dl1=0, dL/dl2=0, dL/dl3=0. where dL/dx means the partial derivative of L w.r.t. x.
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