Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.7 - Maximum and Minimum Values - 14.7 Exercise - Page 968: 32


Absolute minimum: $f(0,0)=0$ Absolute maximum: $f(4,0)=4$

Work Step by Step

To calculate the critical points, we have to first put $f_x(x,y), f_y(x,y)$ equal to $0$. Thus, $f_x=1-y,f_y=1-x$ This yields, $x=1,y=1$ and $f(1,1)=1$ Also, Thus, $f_{xx}=f_{yy}=0$ This yields, $x=0,y=0$ From the vertices of the triangle, after simplifications, we have: $f(x,y)=x+y-xy$ $=x+(-\dfrac{1}{2}x+2)-x(-\dfrac{1}{2}x+2)$ $=x-\dfrac{1}{2}x+2+\dfrac{1}{2}x^2-2x$ $=\dfrac{1}{2}x^2-\dfrac{3}{2}x+2$ Therefore, we have minimum and maximum values at $x=0,4$ Hence, Absolute minimum: $f(0,0)=0$ Absolute maximum: $f(4,0)=4$
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