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 February 24th, 2011, 04:59 PM #1 Newbie   Joined: Feb 2011 Posts: 21 Thanks: 0 Need Help! Linear approximation! The demand function for a product is given by p=f(q)=70? squareroot( q) where p is the price per unit in dollars for q units. Use the linear approximation to approximate the the price when 1224 units are demanded. Solution: We want to approximate f(1224) . From f(q)?L(q)=f(a)+f'(a)(q?a) and the fact that f'(a)= (choices are ( -1/sqrt(a) ,1/(2sqrt(a)), -1/(2sqrt(a)), 1/sqrt(a)) we choose a= From f(1225)= and f'(1225)= we get f(1224) . Hence, the price per unit when 1224 units are demanded is approximately \$ .
 February 24th, 2011, 05:25 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Need Help! Liner approximation! We may use the linear approximation to f(q) by the point-slope formula: $f(q)\approx L(q)=f#39;(a)(q-a)+f(a)$ With $f(q)=70-\sqrt{q}$ we have $f'(q)=-\frac{1}{2}q^{-\frac{1}{2}}=-\frac{1}{2\sqrt{q}}$ so $f'(a)=-\frac{1}{2\sqrt{a}}$ and since $35^2=1225$ and $1224\approx1225$ we choose a = 1225. Thus with: $f(1225)=70-\sqrt{1225}=70-35=35$ and $f'(1225)=-\frac{1}{2\sqrt{1225}}=-\frac{1}{2\cdot35}=-\frac{1}{70}$ we have: $f(1224)\approx-\frac{1}{70}(1224-1225)+35=\frac{2451}{70}\approx35.0143$ This is very close to the actual value of f(1224).
 February 24th, 2011, 05:43 PM #3 Newbie   Joined: Feb 2011 Posts: 21 Thanks: 0 Re: Need Help! Linear approximation! THANK YOU !!!!

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