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Math Help - Langrangean Method for a maximum

  1. #1
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    Langrangean Method for a maximum

    Hi
    I am very stuck on this part of a question. The first part was easy enough, and basically I found out the following information about a farmer: The farmer has £ 15,000 to spend on renting land A, and onfertiliser F. I worked out previously that the the rent for an acre of land is £ 10 and thatFertiliser costs £ 100 per ton. If the
    farmer’s yield in bushels of wheat is
    W(A, F) = 400AF^0.5


    by using the Lagrangean method, fi…nd the amounts of land and fertiliser which will
    maximise the farmer’s yield. Verify that you have a maximum. How would you
    measure the marginal increase in yield of increasing the budget, and what value
    would you obtain?

    Now, I know that I take the partial derivativea and set them equal to zero, but I just don't know where to start. Can anyone help me?

    Thanks
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  2. #2
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    Re: Langrangean Method for a maximum

    Now I think I worked out correctly land A at 1000 and f fertiliser at 50. Can anyone corroborate this and help me with the next part? Thanks
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  3. #3
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    Re: Langrangean Method for a maximum

    You want to maximize W(A, F)= AF^{1/2} with constraint V(A, F)= 10A+ 100F= 15000 (I am assuming that A is the number of acres of land and F is number oftons of fertilizer. That should have been said.)

    The Lagrange multiplier method (I tend to think of the "Lagrangean" as being a physica function) says that at a max or min we will have \nabla W= <F^{1/2},(1/2)AF^{-1/2}>= \lambda\nabla V= \lambda<10, 100> which gives the two equations F^{1/2}= 10\lambda, (1/2)AF^{-1/2}= 100\lambda.
    Since a specific value of \lambda is not part of the solution, I find it simplest, often, to eliminate \lambda by dividing one equation by the other: \frac{F^{1/2}}{(1/2)AF^{-1/2}}= \frac{2F}{A}= \frac{10}{100}=\frac{1}{10}. That is, A= 20F. Putting that into the constraint, 10A+ 100F= 200F+ 100F= 300F= 15000 and so F= 50 and then A= 20(50)= 1000, exactly as you say.

    Well done!

    The "marginal increase" of increasing the budget, b, would be \frac{dW}{db}= \frac{\partial W}{\partial A}\frac{dA}{db}+ \frac{\partial W}{\partial F}\frac{dF}{db} and, because W= 400AF^{1/2}, \frac{\partial W}{\partial A}= 400F^{1/2} and \frac{\partial W}{\partial F}= 200AF^{-1/2}.
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