Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.7 Maximum and Minimum Values - 14.7 Exercises - Page 1008: 31


Absolute minimum: $f(1,0)=-1$ Absolute maximum: $f(0,2)=f(0,-2)=4$

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, $f_x=2x-2,f_y=2y$ This yields, $x=1,y=0$ and $f(0,1)=-1$ From the vertices of the triangle, after simplifications, we have $x=x ,y=2-x,x-2$ Now, $f(x,2-x)=x^2+(2-x)^2-2x=2x^2-6x+4$ and $f(x,x-2)=x^2+(x-2)^2-2x=2x^2-6x+4$ Also, $f'(x,2-x)=4x-6$ and $f'(x,x-2)=4x+6$ This gives $x=\dfrac{3}{2},\dfrac{3}{2}$,$y=\dfrac{1}{2},-\dfrac{1}{2}$ Hence, Absolute minimum: $f(1,0)=-1$ Absolute maximum: $f(0,2)=f(0,-2)=4$
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