Calculus (3rd Edition)

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

Chapter 15 - Differentiation in Several Variables - 15.7 Optimization in Several Variables - Exercises - Page 822: 35

Answer

(a) There is only one critical point at $\left( {\frac{1}{3},\frac{1}{3}} \right)$. The extreme value is $f\left( {\frac{1}{3},\frac{1}{3}} \right) = \frac{1}{3}$. (b) The extreme values of $f$ on the bottom edge is $g\left( {\frac{1}{2}} \right) = f\left( {\frac{1}{2},0} \right) = \frac{1}{4}$. (c) 1. On the top edge of the square, $y=2$. The extreme value of $f$ along $y=2$ is $f\left( { - \frac{1}{2},2} \right) = - \frac{7}{4}$. 2. On the left edge of the square, $x=0$. The extreme value of $f$ along $x=0$ is $f\left( {0,\frac{1}{2}} \right) = \frac{1}{4}$. 3. On the right edge of the square, $x=2$. The extreme value of $f$ along $x=2$ is $f\left( {2, - \frac{1}{2}} \right) = - \frac{7}{4}$. (d) The largest value of $f$ among the values computed in (a), (b), and (c) is $f\left( {\frac{1}{3},\frac{1}{3}} \right) = \frac{1}{3}$. The maximum of $f$ is $\frac{1}{3}$ which occurs at the critical point $\left( {\frac{1}{3},\frac{1}{3}} \right)$.

Work Step by Step

(a) We have $f\left( {x,y} \right) = x + y - {x^2} - {y^2} - xy$. The partial derivatives are ${f_x} = 1 - 2x - y$, ${\ \ }$ ${f_y} = 1 - 2y - x$ To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$: ${f_x} = 1 - 2x - y = 0$, ${\ \ }$ ${f_y} = 1 - 2y - x = 0$ From the first equation we get $y=1-2x$. Substituting it in the second equation gives $1 - 2 + 4x - x = 0$ So, the solutions are $x = \frac{1}{3}$ and $y = \frac{1}{3}$. There is only one critical point at $\left( {\frac{1}{3},\frac{1}{3}} \right)$. So, the extreme value $f\left( {\frac{1}{3},\frac{1}{3}} \right) = \frac{1}{3}$. (b) On the bottom edge of the square, $y=0$. Write $g\left( x \right) = f\left( {x,0} \right) = x - {x^2}$. Find the extreme values of $f$ on the bottom edge. The critical points of $g$ can be found by solving the equation $g'\left( x \right) = 0$. $g'\left( x \right) = 1 - 2x = 0$ $x = \frac{1}{2}$ So, the extreme values of $f$ on the bottom edge is $g\left( {\frac{1}{2}} \right) = f\left( {\frac{1}{2},0} \right) = \frac{1}{4}$. (c) 1. On the top edge of the square, $y=2$. Write $h\left( x \right) = f\left( {x,2} \right) = - 2 - {x^2} - x$. Find the extreme values of $f$ on the top edge. The critical points of $h$ can be found by solving the equation $h'\left( x \right) = 0$. $h'\left( x \right) = - 2x - 1 = 0$ $x = - \frac{1}{2}$ So, the extreme value of $f$ along $y=2$ is $h\left( { - \frac{1}{2}} \right) = f\left( { - \frac{1}{2},2} \right) = - \frac{7}{4}$. 2. On the left edge of the square, $x=0$. Write $m\left( y \right) = f\left( {0,y} \right) = y - {y^2}$. Find the extreme values of $f$ on the left edge. The critical points of $m$ can be found by solving the equation $m'\left( y \right) = 0$. $m'\left( y \right) = 1 - 2y = 0$ $y = \frac{1}{2}$ So, the extreme value of $f$ along $x=0$ is $m\left( {\frac{1}{2}} \right) = f\left( {0,\frac{1}{2}} \right) = \frac{1}{4}$. 3. On the right edge of the square, $x=2$. Write $n\left( y \right) = f\left( {2,y} \right) = - 2 - y - {y^2}$. Find the extreme values of $f$ on the right edge. The critical points of $n$ can be found by solving the equation $n'\left( y \right) = 0$. $n'\left( y \right) = - 1 - 2y = 0$ $y = - \frac{1}{2}$ So, the extreme value of $f$ along $x=2$ is $n\left( { - \frac{1}{2}} \right) = f\left( {2, - \frac{1}{2}} \right) = - \frac{7}{4}$. (d) The largest value of $f$ among the values computed in (a), (b), and (c) is $f\left( {\frac{1}{3},\frac{1}{3}} \right) = \frac{1}{3}$. Hence, the maximum of $f$ is $\frac{1}{3}$ which occurs at the critical point $\left( {\frac{1}{3},\frac{1}{3}} \right)$.
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