[SOLVED] Economics Lagrangean Method Problem

SOLVED

Quote:

A firm has decided through regression analysis that its sales (S) are a function of

the amount of advertising (measured in units) in two different media, television (x) and magazines (y):

S(x, y) = 100 – x^{2} + 30x – y^{2} + 40y

** (a)** Find the level of TV and magazine advertising units that maximizes the firm's sales.

** (b)** Suppose that the advertising budget is restricted to 31 units. Determine the level of advertising (in units) that maximizes sales subject to this budget constraint.

I already solved (a) by finding the derivative with respect to (x,y) and equating to 0.

(a) X^{*} = 15, Y^{*} = 20

My teacher provided the answer to (b) and I have absolutely no idea to how he arrived at it. I assumed the the Income(M)=31 but without any given prices for (x,y), I cannot seem to apply it into a Lagrangean method. Could anyone help me?

(b) X^{*} = 13, Y^{*} = 18

Re: Economics Lagrangean Method Problem

What, exactly, have you tried? It seems pretty straight forward to me. The problem is to maximize $\displaystyle S(x, y)= 100- x^2+ 30x- y^2+ 40x$ with the constraint that f(x,y)= x+ y= 31.

Lagrange's method says that, at a critical point, $\displaystyle \nabla S= \lambda\nabla f$.

That is, $\displaystyle <-2x+ 30, -2y+ 40>= \lambda <1, 1>$ so we have the two equations $\displaystyle -2x+ 30= \lambda$ and $\displaystyle -2y+ 40= \lambda$.

Re: Economics Lagrangean Method Problem

Thanks.

I just realized that I had been so used to my instructor's problems where he provides the prices for each product that I had no idea what to do when he doesn't give any prices.

(It works on problems which are similar to the one I just asked for.)

I actually spent hours trying to find a hidden price in the question somewhere.

:<