Date: 10/15/2003 at 05:00:24 From: Matthew Subject: Optimisation A firm with a total income of $50,000 per year currently spends $10,000 per year on advertising. It finds that by doubling advertising expenditure, total income increases by twenty percent. By maximising the total residue of income less advertising expediture, find the optimum level of advertising for the firm. (Hint: d/dx a^x = a^x ln a.) I don't understand where to start the question.
Date: 10/15/2003 at 07:55:28 From: Doctor Luis Subject: Re: Optimisation Hi Matthew, Currently, the firm has a total income of $50,000/year, and it spends $10,000/year on advertising, leaving us with a residue profit of $40,000/year. Now, if advertising was doubled to $20,000/year, its income would increase 20% to $60,000/year. The residue from this would still be $40,000/year. Since this gives us the same net result, you gain nothing from having spent all that money doubling the advertising expenditure. But is there a happy medium? Assuming that every doubling of advertising leads to 20% growth in income, the residue R(x) as a function of the level of advertising x is R(x) = ($50,000/year)*(1.20)^x - ($10,000/year)*(2)^x where the advertising grows as A(x) = ($10,000/year)*(2)^x and the income grows as I(x) = ($50,000/year)*(1.20)^x Thus, we are comparing the growth of two exponential functions. 2^x grows faster than 1.2^x, but the income started with a higher coefficient. Eventually, A(0)*2^x catches up to I(0)*1.2^x and it becomes counterproductive to keep increasing the advertising budget. We have already seen R(0)=$40,000/yr, and R(1)=$40,000/yr, so we can guess that by the time x=1, most of the gains from advertising have been cancelled by the cost of advertising itself. Therefore, our happy medium is somewhere between x=0 and x=1. Here's the plot of R(x), Clearly, the residue profit maxes out somewhere in between advertising levels x = 0 and x = 1, just as we expected. Now you can just use calculus to find the optimal value of x. I hope this helped! Let us know if you have any more questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/
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