calc help....please! - Zilvia.net Forums | Nissan 240SX (Silvia) and Z (Fairlady) Car Forum

withy · 11-06-2006, 01:00 PM

i'm working on some on my calc homework and i'm having a little trouble with this problem in particular:
the demand function is modeled by:

P = 740-x^1/2 , 0 <=x <= 740

find the interval on which the demand is elastic, inelastic, and of unit elasticity.

price elasticity = (p/x)/(dp/dx)

so i know that i have to substitute (740-x)^1/2 into the equation..... but how?

i e-mailed the math lab hear and they gave me this:

(((740-x)^1/2)/x)/(-1/(2/((740-x)^1/2)))

so my question to everyone is why is d/dx = (-1/(2/((740-x)^1/2)))?!?!?!?!?

its a simple problem but i jsut can't see it....

projekt_s13 · 11-06-2006, 01:23 PM

ummmmm........... yea......... Sorry i cant help you there. o_O

withy · 11-06-2006, 01:42 PM

ehhh, thanks for looking.

*ManoNegra* · 11-06-2006, 01:56 PM

P is a function of x so -> p(x) = 740 - x^1/2

derivative of p(x) = p'(x) = -(1/2)x^(-1/2) = -1/[2x^1/2]

p/x = (740 - x^1/2) / x

so (p/x)/(dp/dx) = [(740 - x^1/2) / x ] / [-1/(2x^1/2)]

simplify to get this:

price elasticity = 1/2 - 370/x^1/2 = e(x) for simplicity

e'(x) = 160 / x^3/2 derivative of price elasticity function

You can graph this function to find out the interval of growth (elasticity?), decline (inelastic?) and unity (equate the function to 1 and solve for x).

Been a while so make sure you double check your work with a classmate or teacher.

11-06-2006, 01:00 PM	#1
withy Leaky Injector Join Date: May 2005 Location: wichita, ks Age: 38 Posts: 120 Trader Rating: (0) Feedback Score: 0 reviews	calc help....please! i'm working on some on my calc homework and i'm having a little trouble with this problem in particular: the demand function is modeled by: P = 740-x^1/2 , 0 <=x <= 740 find the interval on which the demand is elastic, inelastic, and of unit elasticity. price elasticity = (p/x)/(dp/dx) so i know that i have to substitute (740-x)^1/2 into the equation..... but how? i e-mailed the math lab hear and they gave me this: (((740-x)^1/2)/x)/(-1/(2/((740-x)^1/2))) so my question to everyone is why is d/dx = (-1/(2/((740-x)^1/2)))?!?!?!?!? its a simple problem but i jsut can't see it....

11-06-2006, 01:23 PM	#2
projekt_s13 Post Whore! Join Date: Jul 2005 Location: Chico/Pittsburg, CA Age: 38 Posts: 2,801 Trader Rating: (5) Rides Gallery: projekt_s13 Feedback Score: 5 reviews	ummmmm........... yea......... Sorry i cant help you there. o_O __________________

11-06-2006, 01:42 PM	#3
withy Leaky Injector Join Date: May 2005 Location: wichita, ks Age: 38 Posts: 120 Trader Rating: (0) Feedback Score: 0 reviews	ehhh, thanks for looking.

11-06-2006, 01:56 PM	#4
ManoNegra Guild of Skullduggerous Intent Join Date: Jan 2004 Location: Socal Posts: 7,841 Trader Rating: (48) Feedback Score: 48 reviews	P is a function of x so -> p(x) = 740 - x^1/2 derivative of p(x) = p'(x) = -(1/2)x^(-1/2) = -1/[2x^1/2] p/x = (740 - x^1/2) / x so (p/x)/(dp/dx) = [(740 - x^1/2) / x ] / [-1/(2x^1/2)] simplify to get this: price elasticity = 1/2 - 370/x^1/2 = e(x) for simplicity e'(x) = 160 / x^3/2 derivative of price elasticity function You can graph this function to find out the interval of growth (elasticity?), decline (inelastic?) and unity (equate the function to 1 and solve for x). Been a while so make sure you double check your work with a classmate or teacher.