I am studying for my exams and got very very stuck at a word problem on the Lagrange Methods, my biggest difficulty is to properly identify the function to be maximized (in this case) and so its constraint.

Problem. "A computer company has a monthly advertising budget of 20.000 dolars. its marketing department estimates that if x dolars are spent each month on advertising in newspapers and y dollars per month on advertising in television, then the monthy sales will be given as $$S = 80x^{1/4} y^{3/4}$$ dollars. If the profit is 10% of sales less the advertising cost, determine how to allocate the advertising budget in order to maximize the monthly profit."

  • should I assume that the variables x and y from the cost function $$x+y= 20.000$$ are the same as the ones from sales? I would think that sales is a function in terms of quantity of product not of advertising, but the solution seens so simplistic that I tried as so.
  • I did then $$L(x,y,\lambda) = (80x^{1/4} y^{3/4})*0.1 - 20.000 - \lambda *(x+y-20.000)$$
  • I also tried substituting the cost function in terms of one variable into the "Profit" but nothing seens to work. Book solution is given as x = 5.000 and y = 15.000

Thank you for any tip on how to find equations in word problem, this being an example.

  • $\begingroup$ Yes, I believe your "Lagrange function" is set up correctly. You should get $$ \lambda \ = \ 2 \ \frac{y^{3/4}}{x^{3/4}} \ = \ 6 \ \frac{x^{1/4}}{y^{1/4}} \ \ , $$ which will give you a simple relation between the values of $ \ x \ $ and $ \ y \ $ after a little algebra. $\endgroup$ – colormegone May 23 '16 at 8:55
  • $\begingroup$ Yes I think the variables are the same as the ones from sales. Also, it does make sense for sales to be an increasing function of the amount of resources put into advertising as adverts drive consumer demand. It seems likely (given the physical situation) that you will want to use the full 20000 budget in spending. But, if you aren't happy just going with the flow, you can use the usual methods to find any local critical points (of the profit function ) in the region bounded by $x=0$, $y=0$ and $x+y =20000$. You can then compare these with the maximum along $x+y=20000$. $\endgroup$ – Josh R May 23 '16 at 9:03
  • $\begingroup$ As to your second question, using a profit function $$ P(x) \ = \ 8 \ \cdot \ x^{1/4} \ \cdot \ (20000 - x)^{3/4} \ - \ 20000 \ \ , $$ and setting the derivative equal to zero to find the critical value for $ \ x \ $ does work, but you will need to be careful with the Product Rule and Chain Rule [in particular, don't forget that you pick up a minus-sign in differentiating $ \ (20000 - x ) \ $ ] . $\endgroup$ – colormegone May 23 '16 at 9:07
  • $\begingroup$ Thank you so much Reckless, it was easier than I thought, your confirmations just helped me to find a solution. Many thanks. $\endgroup$ – DAguilera May 23 '16 at 11:51

Your cost function can be simply

$$f(x, y) = x^{1/4} y^{3/4}$$

since the profit,

$$p(x, y) = 20 - 8 f(x, y)$$ (that is 10% of sales less than the advert cost of 20).

Formulate the problem as

$$\arg \underset{x, y}{\max} \, x^{1/4} y^{3/4}$$

Subject to: $x + y = 20$.

Therefore, $$L(x, y, \lambda) = x^{1/4} y^{3/4} - \lambda (x + y - 20)$$

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