Linear approximation of bus revenue

bcahmel
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Homework Statement


If the price of a bus pass from Albuquerque to Los Alamos is set at x dollars, a bus company takes in a monthly revenue of R(x) = 1.5x − 0.01x2 (in thousands of dollars).

Suppose that x = 80. How will revenue be affected by a small increase in price? Explain using the Linear Approximation.


The Attempt at a Solution


first I took the derivative which is 1.5-0.02x. Then I plugged 80 in for x, so f'(x)=-0.1. Is this right? So the revenue will slightly decrease.
 
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so the linear approximation is
R(x) \approx R(x_0) +R'(x_0)(x - x_0)
 
ok, so its just like y=mx+b, sort of..
 
Which is why they call it a linear approximation...
 
yes, I get it now...thanks mark44
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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