Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 19: 43


$ y= \frac{1}{3}$ is the minimum, when $ x=\frac{1}{6}$

Work Step by Step

Given $$ y=4x-12x^{2} $$ So, we have \begin{aligned} y&=4x-12x^{2} \\ &=-12\left(x^{2}-\frac{1}{3} x\right)\\ &=-12\left(x^{2}-\frac{1}{3} x+\frac{1}{36}-\frac{1}{36}\right)\\ &=-12\left((x-\frac{1}{6} )-\frac{1}{36}\right)\\ &=-12\left((x-\frac{1}{6} )^2-\frac{1}{36}\right)\\ \end{aligned} The lowest value for the squared term is $0$. So, we see that $ y=\frac{12}{36}=\frac{1}{3}$ is the minimum, when $ x=\frac{1}{6}$.
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