Linear Approximation: Demand

Tulki · #1 $Old$ November 2nd, 2009, 08:16 PM

The demand function for a product is given by:

$p = f(q) = 90 - sqrt(q)$

where p is the price per unit in dollars for q units. Use the linear approximation to approximate the price when 2024 units are demanded.

We want to approximate f(2024). From $f(q) ~= L(q) = f(a) + f'(a)(q-a)$

And the fact that $f'(a) = -1/(2sqrt(a))$ , we choose 2025 for a.

from f(2025) = 45 and f'(2025) = -22.5 we get f(2024) ~= 67.50

Hence, the price per unit when 2024 units are demanded is approximately ______.

Did I mess up anywhere here? And how do I find the final answer? Totally new to linear approximation and I'm trying to get a bit of a head start on my own. Thanks!

HallsofIvy · #2 $Old$ November 3rd, 2009, 05:04 AM

Quote:

Originally Posted by Tulki $View Post$

The demand function for a product is given by:

$p = f(q) = 90 - sqrt(q)$

where p is the price per unit in dollars for q units. Use the linear approximation to approximate the price when 2024 units are demanded.

We want to approximate f(2024). From $f(q) ~= L(q) = f(a) + f'(a)(q-a)$

And the fact that $f'(a) = -1/(2sqrt(a))$ , we choose 2025 for a.

Good!

Quote:

from f(2025) = 45 and f'(2025) = -22.5 we get f(2024) ~= 67.50

You give the formula for f', correctly, as $f'(a) = -1/(2sqrt(a))$ but you appear to have calculated $-\sqrt{2025)}/2$ !
$\sqrt{2025}$ is 45 so [math]-1/(2\sqrt{2025})= -1/90= -0.01111. Since p has decreased by 1, to go from f2025) to f(2024) we need to subtract the derivative, as you did, but 45- (-0.01111)= 44.98888 or, since this is a price in dollars, $44.99.

Quote:

Hence, the price per unit when 2024 units are demanded is approximately ______.

Did I mess up anywhere here? And how do I find the final answer? Totally new to linear approximation and I'm trying to get a bit of a head start on my own. Thanks!

Tulki · #3 $Old$ November 4th, 2009, 05:31 PM

Sorry about the late response. Thank you very much though!