# Thread: Economics Lagrangean Method Problem

1. ## [SOLVED] Economics Lagrangean Method Problem

SOLVED

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 – x2 + 30x – y2 + 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

2. ## Re: Economics Lagrangean Method Problem

What, exactly, have you tried? It seems pretty straight forward to me. The problem is to maximize $S(x, y)= 100- x^2+ 30x- y^2+ 40x$ with the constraint that f(x,y)= x+ y= 31.
Lagrange's method says that, at a critical point, $\nabla S= \lambda\nabla f$.
That is, $<-2x+ 30, -2y+ 40>= \lambda <1, 1>$ so we have the two equations $-2x+ 30= \lambda$ and $-2y+ 40= \lambda$.

3. ## 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.
:<