# Math Help - Langrangean Method for a maximum

1. ## 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

2. ## 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

3. ## 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 $\nabla W= = \lambda\nabla V= \lambda<10, 100>$ 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 $\frac{\partial W}{\partial F}= 200AF^{-1/2}$.