Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 1 - Linear Functions - 1.2 Linear Functions and Applications - 1.2 Exercises - Page 24: 31



Work Step by Step

The function is $p=S(q)=0.3q+2.7$ with $D(2)=6.1$ which is also $(q_2,p_2)=(2,6.1)$ To find the quantity demanded at a price of $\$4.50$ per watermelon, replace p in the demand function with $4.50$ and solve for q. $4.50=0.3q+2.7$ $0.3q=1.8$ $q=6$ To find m we use the formula: $m=\frac{p_2-p_1}{q_2-q_1}=\frac{6.1-4.5}{2-6}=-\frac{4}{5}=-0.4$ Assume that the demand function is linear, we have $p-p_1=m(q-q_1)$ $D(q)-4.5=-0.4(q-q_1)$ $D(q)-4.5=-0.4q+2.4$ $D(q)=-0.4q+6.9$
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